MATH-181 (Calculus I) Curriculum

Session 09 - Implicit Differentiation and Logarithmic Differentiation

Handout Worksheet
Fall Semester 2026
Session Objectives & Overview

By the end of this session, you will be able to:

  • Formulate and calculate first derivatives (dydx\frac{dy}{dx}dxdy​) of implicit mathematical and physical constraints using the Chain Rule, product rule, and algebraic isolation techniques.
  • Execute logarithmic differentiation to calculate derivatives of complex multi-factor rational functions and variable-base variable-exponent functions (y=[f(x)]g(x)y = [f(x)]^{g(x)}y=[f(x)]g(x)), isolating fractional and relative rates of change.
  • Compute higher-order implicit derivatives (d2ydx2\frac{d^2y}{dx^2}dx2d2y​) through nested quotient rules, back-substituting the first derivative, and simplifying expressions using the original curve constraint equations to evaluate geometric curvature.

Micro-Lecture

1. Implicit Relations & Hidden Physical Constraints

  • Mathematical Foundation: Up to this point, curves have been written in explicit form, y=f(x)y = f(x)y=f(x), where yyy is isolated entirely on one side. However, physical constraints often bind variables implicitly, such as circles (x2+y2=r2x^2 + y^2 = r^2x2+y2=r2) or closed kinematic loops (x2y−2=3y3x^2y - 2 = 3y^3x2y−2=3y3). These curves often fail the Vertical Line Test, yet every localized smooth segment behaves as a differentiable function y(x)y(x)y(x). The foundational principle of implicit differentiation is treating yyy as an implicit function of xxx and applying the Chain Rule whenever differentiating a yyy term:
    ddx[g(y)]=g′(y)⋅dydx\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}dxd​[g(y)]=g′(y)⋅dxdy​
    (see Implicit Differentiation).
  • Engineering Context: Mechanical and robotics engineers model closed linkage chains, four-bar mechanisms, and automated cutting paths along elliptical rails. Thermodynamics engineers rely on implicit equations of state (such as the Van der Waals equation) to calculate the instantaneous rate of pressure change (dPdV\frac{dP}{dV}dVdP​) during gas expansion.

2. Logarithmic Differentiation: Taming Towers and Multi-Factor Systems

  • Mathematical Foundation: Calculus provides the Power Rule for constant powers (ddx[xn]=nxn−1\frac{d}{dx}[x^n] = n x^{n-1}dxd​[xn]=nxn−1) and the Exponential Rule for constant bases (ddx[bx]=ln⁡(b)bx\frac{d}{dx}[b^x] = \ln(b) b^xdxd​[bx]=ln(b)bx). Neither rule applies to variable bases with variable exponents (y=[f(x)]g(x)y = [f(x)]^{g(x)}y=[f(x)]g(x), like y=xxy = x^xy=xx). Applying ln⁡\lnln to both sides transforms powers into products:
    ln⁡y=g(x)⋅ln⁡(f(x))\ln y = g(x) \cdot \ln(f(x))lny=g(x)⋅ln(f(x))
    Differentiating implicitly gives 1ydydx\frac{1}{y}\frac{dy}{dx}y1​dxdy​, converting an impossible exponential derivative into an elementary Product Rule problem (see Logarithmic Differentiation).
  • Engineering Context: In electrical engineering, transmission line signal decay models intertwine exponential decay, sinusoidal oscillations, and radical damping factors. Logarithmic differentiation bypasses nested quotient rules to instantly calculate signal attenuation loss. In materials science, the logarithmic derivative directly yields the fractional rate of thermal expansion (1vdvdT\frac{1}{v}\frac{dv}{dT}v1​dTdv​).

3. Higher-Order Implicit Rates & Curvature Analysis

  • Mathematical Foundation: While the first derivative dydx\frac{dy}{dx}dxdy​ provides the tangent slope, the second derivative d2ydx2\frac{d^2y}{dx^2}dx2d2y​ quantifies concavity and physical curvature. Differentiating dydx\frac{dy}{dx}dxdy​ with respect to xxx triggers the Quotient Rule and introduces new dydx\frac{dy}{dx}dxdy​ terms. A complete solution requires back-substituting the first derivative and simplifying using the curve's original constraint equation (see Higher-Order Implicit Differentiation).
  • Engineering Context: Aerospace engineers compute the second derivative along circular viewport bulkheads to locate thermal stress concentrations during atmospheric re-entry. Automotive chassis engineers analyze d2ydx2\frac{d^2y}{dx^2}dx2d2y​ along suspension geometry curves to control dynamic wheel camber angles and maintain tire grip during cornering.

Skill Block 1

Worked Example A: Differentiating Basic Polynomial Relations with Product Rule

Find dydx\frac{dy}{dx}dxdy​ for the implicit relation x2y−2=3y3x^2y - 2 = 3y^3x2y−2=3y3.

Worked Solution:
  1. Governing Formula: The Product Rule and Implicit Chain Rule:
    ddx[uv]=dudxv+udvdx,ddx[g(y)]=g′(y)dydx\frac{d}{dx}[uv] = \frac{du}{dx}v + u\frac{dv}{dx}, \qquad \frac{d}{dx}[g(y)] = g'(y)\frac{dy}{dx}dxd​[uv]=dxdu​v+udxdv​,dxd​[g(y)]=g′(y)dxdy​
    (These rules are defined in Product Rule and Implicit Differentiation.)
  2. Differentiate both sides with respect to xxx:
    ddx[x2y]−ddx[2]=ddx[3y3]\frac{d}{dx}[x^2y] - \frac{d}{dx}[2] = \frac{d}{dx}[3y^3]dxd​[x2y]−dxd​[2]=dxd​[3y3]
    (2x⋅y+x2dydx)−0=9y2dydx\left(2x \cdot y + x^2 \frac{dy}{dx}\right) - 0 = 9y^2 \frac{dy}{dx}(2x⋅y+x2dxdy​)−0=9y2dxdy​
    2xy+x2dydx=9y2dydx2xy + x^2\frac{dy}{dx} = 9y^2\frac{dy}{dx}2xy+x2dxdy​=9y2dxdy​
  3. Collect terms containing dydx\frac{dy}{dx}dxdy​ on one side:
    2xy=9y2dydx−x2dydx2xy = 9y^2\frac{dy}{dx} - x^2\frac{dy}{dx}2xy=9y2dxdy​−x2dxdy​
  4. Factor out dydx\frac{dy}{dx}dxdy​ and solve:
    2xy=dydx(9y2−x2)  ⟹  dydx=2xy9y2−x22xy = \frac{dy}{dx}(9y^2 - x^2) \implies \frac{dy}{dx} = \frac{2xy}{9y^2 - x^2}2xy=dxdy​(9y2−x2)⟹dxdy​=9y2−x22xy​

Worked Example B: Managing Trigonometric Chains in Implicit Relations

Find dydx\frac{dy}{dx}dxdy​ for the equation cos⁡(y)=x2−y2\cos(y) = x^2 - y^2cos(y)=x2−y2.

Worked Solution:
  1. Governing Formula: The Trigonometric Chain Rule:
    ddx[cos⁡(y)]=−sin⁡(y)dydx\frac{d}{dx}[\cos(y)] = -\sin(y)\frac{dy}{dx}dxd​[cos(y)]=−sin(y)dxdy​
    (This rule is defined in Implicit Differentiation.)
  2. Differentiate both sides with respect to xxx:
    −sin⁡(y)dydx=2x−2ydydx-\sin(y)\frac{dy}{dx} = 2x - 2y\frac{dy}{dx}−sin(y)dxdy​=2x−2ydxdy​
  3. Collect terms containing dydx\frac{dy}{dx}dxdy​ on the left side:
    2ydydx−sin⁡(y)dydx=2x2y\frac{dy}{dx} - \sin(y)\frac{dy}{dx} = 2x2ydxdy​−sin(y)dxdy​=2x
  4. Factor out dydx\frac{dy}{dx}dxdy​ and solve:
    dydx(2y−sin⁡y)=2x  ⟹  dydx=2x2y−sin⁡y\frac{dy}{dx}(2y - \sin y) = 2x \implies \frac{dy}{dx} = \frac{2x}{2y - \sin y}dxdy​(2y−siny)=2x⟹dxdy​=2y−siny2x​

Worked Example C: Solving Linear Systems in dydx\frac{dy}{dx}dxdy​

Find dydx\frac{dy}{dx}dxdy​ for the relation y2−2x=1−2yy^2 - 2x = 1 - 2yy2−2x=1−2y.

Worked Solution:
  1. Governing Formula: The Power Chain Rule:
    ddx[yn]=nyn−1dydx\frac{d}{dx}[y^n] = n y^{n-1}\frac{dy}{dx}dxd​[yn]=nyn−1dxdy​
    (This rule is defined in Implicit Differentiation.)
  2. Differentiate both sides with respect to xxx:
    2ydydx−2=0−2dydx2y\frac{dy}{dx} - 2 = 0 - 2\frac{dy}{dx}2ydxdy​−2=0−2dxdy​
  3. Collect dydx\frac{dy}{dx}dxdy​ terms on the left side and constants on the right side:
    2ydydx+2dydx=22y\frac{dy}{dx} + 2\frac{dy}{dx} = 22ydxdy​+2dxdy​=2
  4. Factor out dydx\frac{dy}{dx}dxdy​ and simplify:
    dydx(2y+2)=2  ⟹  dydx=22(y+1)=1y+1\frac{dy}{dx}(2y + 2) = 2 \implies \frac{dy}{dx} = \frac{2}{2(y + 1)} = \frac{1}{y + 1}dxdy​(2y+2)=2⟹dxdy​=2(y+1)2​=y+11​

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

  • (a) In the Conceptual Framework, explain why differentiating an expression with respect to xxx treats x3x^3x3 and y3y^3y3 differently. What role does the Chain Rule play?
  • (b) Deconstruct Worked Example B: Why does the left-side derivative contain −sin⁡(y)dydx-\sin(y)\frac{dy}{dx}−sin(y)dxdy​ rather than simply −sin⁡(y)-\sin(y)−sin(y)? What common error does this prevent?
  • (c) Review the linked Engineering Applications (Mechanical Kinematics (Robotic Linkages)). Why must the industrial cutting robot compute the normal slope mnormal=−1dy/dxm_{\text{normal}} = -\frac{1}{dy/dx}mnormal​=−dy/dx1​ from the implicit ellipse rather than relying on an explicit function?
Solution Key:
  • (a) In single-variable calculus with respect to xxx, the independent variable changes freely, so ddx[x3]=3x2⋅dxdx=3x2(1)=3x2\frac{d}{dx}[x^3] = 3x^2 \cdot \frac{dx}{dx} = 3x^2(1) = 3x^2dxd​[x3]=3x2⋅dxdx​=3x2(1)=3x2. However, yyy is dependent on xxx through the relation (y=y(x)y = y(x)y=y(x)). By the Chain Rule, differentiating the outer layer with respect to yyy must be chained with the instantaneous rate of change of the inner function with respect to xxx: ddx[y3]=3y2⋅dydx\frac{d}{dx}[y^3] = 3y^2 \cdot \frac{dy}{dx}dxd​[y3]=3y2⋅dxdy​.
  • (b) In Worked Example B, the term cos⁡(y)\cos(y)cos(y) represents a composite function where the outer shell is cos⁡(u)\cos(u)cos(u) and the inner function is u=y(x)u = y(x)u=y(x). Applying the Chain Rule yields ddx[cos⁡(y)]=−sin⁡(y)⋅dydx\frac{d}{dx}[\cos(y)] = -\sin(y) \cdot \frac{dy}{dx}dxd​[cos(y)]=−sin(y)⋅dxdy​. Forgetting the factor dydx\frac{dy}{dx}dxdy​ is the single most common student error in implicit differentiation, as it leaves no derivative term to isolate.
  • (c) In Mechanical Kinematics (Robotic Linkages), the plasma cutter follows the closed ellipse 4x2+9y2=364x^2 + 9y^2 = 364x2+9y2=36. Because an ellipse has two yyy-values for every xxx-value, solving for y=±239−x2y = \pm \frac{2}{3}\sqrt{9-x^2}y=±32​9−x2​ introduces cumbersome square root branches. Implicit differentiation yields dydx=−4x9y\frac{dy}{dx} = -\frac{4x}{9y}dxdy​=−9y4x​ directly. Evaluating this at (32,2)\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)(2​3​,2​) gives tangent slope −23-\frac{2}{3}−32​, so the automated wrist motor aligns to the normal slope mnormal=32=1.5m_{\text{normal}} = \frac{3}{2} = 1.5mnormal​=23​=1.5 perpendicular to the cut surface.
Show your steps and calculations here.

Practice Problem 2: Mixed Polynomial and Product Rule Implicit Derivative

Find dydx\frac{dy}{dx}dxdy​ for the function y=f(x)y = f(x)y=f(x) implicitly defined by the polynomial relation:

x5+y5=1+xyx^5 + y^5 = 1 + xyx5+y5=1+xy

Solution Key:
  1. Governing Formula: The Product Rule and Implicit Chain Rule:
    ddx[uv]=u′v+uv′,ddx[yn]=nyn−1dydx\frac{d}{dx}[uv] = u'v + uv', \qquad \frac{d}{dx}[y^n] = n y^{n-1}\frac{dy}{dx}dxd​[uv]=u′v+uv′,dxd​[yn]=nyn−1dxdy​
    (These rules are defined in Product Rule and Implicit Differentiation.)
  2. Differentiate both sides with respect to xxx:
    ddx[x5]+ddx[y5]=ddx[1]+ddx[xy]\frac{d}{dx}[x^5] + \frac{d}{dx}[y^5] = \frac{d}{dx}[1] + \frac{d}{dx}[xy]dxd​[x5]+dxd​[y5]=dxd​[1]+dxd​[xy]
    5x4+5y4dydx=0+(1⋅y+xdydx)5x^4 + 5y^4\frac{dy}{dx} = 0 + \left(1 \cdot y + x\frac{dy}{dx}\right)5x4+5y4dxdy​=0+(1⋅y+xdxdy​)
    5x4+5y4dydx=y+xdydx5x^4 + 5y^4\frac{dy}{dx} = y + x\frac{dy}{dx}5x4+5y4dxdy​=y+xdxdy​
  3. Collect terms containing dydx\frac{dy}{dx}dxdy​ on the left side:
    5y4dydx−xdydx=y−5x45y^4\frac{dy}{dx} - x\frac{dy}{dx} = y - 5x^45y4dxdy​−xdxdy​=y−5x4
  4. Factor out dydx\frac{dy}{dx}dxdy​ and solve:
    dydx(5y4−x)=y−5x4  ⟹  dydx=y−5x45y4−x\frac{dy}{dx}(5y^4 - x) = y - 5x^4 \implies \frac{dy}{dx} = \frac{y - 5x^4}{5y^4 - x}dxdy​(5y4−x)=y−5x4⟹dxdy​=5y4−xy−5x4​
Show your steps and calculations here.

Practice Problem 3: Exponential and Product Chain Implicit Derivative

Find dydx\frac{dy}{dx}dxdy​ for the implicit relation:

ex−y=2xye^{x-y} = 2xyex−y=2xy

Solution Key:
  1. Governing Formula: The Exponential Chain Rule and Product Rule:
    ddx[eg(x,y)]=eg(x,y)ddx[g(x,y)],ddx[2xy]=2y+2xdydx\frac{d}{dx}[e^{g(x,y)}] = e^{g(x,y)} \frac{d}{dx}[g(x,y)], \qquad \frac{d}{dx}[2xy] = 2y + 2x\frac{dy}{dx}dxd​[eg(x,y)]=eg(x,y)dxd​[g(x,y)],dxd​[2xy]=2y+2xdxdy​
    (These rules are defined in Derivatives of Exponential Functions and Implicit Differentiation.)
  2. Differentiate both sides with respect to xxx:
    ex−y⋅ddx[x−y]=2(1⋅y+xdydx)e^{x-y} \cdot \frac{d}{dx}[x - y] = 2\left(1 \cdot y + x\frac{dy}{dx}\right)ex−y⋅dxd​[x−y]=2(1⋅y+xdxdy​)
    ex−y(1−dydx)=2y+2xdydxe^{x-y}\left(1 - \frac{dy}{dx}\right) = 2y + 2x\frac{dy}{dx}ex−y(1−dxdy​)=2y+2xdxdy​
  3. Distribute the exponential term:
    ex−y−ex−ydydx=2y+2xdydxe^{x-y} - e^{x-y}\frac{dy}{dx} = 2y + 2x\frac{dy}{dx}ex−y−ex−ydxdy​=2y+2xdxdy​
  4. Group all dydx\frac{dy}{dx}dxdy​ terms on the right side:
    ex−y−2y=2xdydx+ex−ydydxe^{x-y} - 2y = 2x\frac{dy}{dx} + e^{x-y}\frac{dy}{dx}ex−y−2y=2xdxdy​+ex−ydxdy​
    ex−y−2y=dydx(2x+ex−y)e^{x-y} - 2y = \frac{dy}{dx}(2x + e^{x-y})ex−y−2y=dxdy​(2x+ex−y)
  5. Isolate dydx\frac{dy}{dx}dxdy​:
    dydx=ex−y−2y2x+ex−y\frac{dy}{dx} = \frac{e^{x-y} - 2y}{2x + e^{x-y}}dxdy​=2x+ex−yex−y−2y​
Show your steps and calculations here.

Practice Problem 4: Trigonometric Quotient Implicit Derivative

Find dydx\frac{dy}{dx}dxdy​ for the implicit equation:

sin⁡(yx)=2x3−3y2\sin\left(\frac{y}{x}\right) = 2x^3 - 3y^2sin(xy​)=2x3−3y2

Solution Key:
  1. Governing Formula: The Trigonometric Chain Rule and Quotient Rule:
    ddx[sin⁡(u)]=cos⁡(u)dudx,ddx[yx]=dydx⋅x−y⋅1x2\frac{d}{dx}[\sin(u)] = \cos(u)\frac{du}{dx}, \qquad \frac{d}{dx}\left[\frac{y}{x}\right] = \frac{\frac{dy}{dx}\cdot x - y\cdot 1}{x^2}dxd​[sin(u)]=cos(u)dxdu​,dxd​[xy​]=x2dxdy​⋅x−y⋅1​
    (These rules are defined in Quotient Rule and Implicit Differentiation.)
  2. Differentiate both sides with respect to xxx:
    cos⁡(yx)⋅[xdydx−yx2]=6x2−6ydydx\cos\left(\frac{y}{x}\right) \cdot \left[ \frac{x\frac{dy}{dx} - y}{x^2} \right] = 6x^2 - 6y\frac{dy}{dx}cos(xy​)⋅[x2xdxdy​−y​]=6x2−6ydxdy​
  3. Multiply through by x2x^2x2 to eliminate the fraction:
    cos⁡(yx)⋅(xdydx−y)=6x4−6x2ydydx\cos\left(\frac{y}{x}\right) \cdot \left(x\frac{dy}{dx} - y\right) = 6x^4 - 6x^2y\frac{dy}{dx}cos(xy​)⋅(xdxdy​−y)=6x4−6x2ydxdy​
  4. Distribute cosine and gather dydx\frac{dy}{dx}dxdy​ terms on the left:
    xcos⁡(yx)dydx−ycos⁡(yx)=6x4−6x2ydydxx\cos\left(\frac{y}{x}\right)\frac{dy}{dx} - y\cos\left(\frac{y}{x}\right) = 6x^4 - 6x^2y\frac{dy}{dx}xcos(xy​)dxdy​−ycos(xy​)=6x4−6x2ydxdy​
    xcos⁡(yx)dydx+6x2ydydx=6x4+ycos⁡(yx)x\cos\left(\frac{y}{x}\right)\frac{dy}{dx} + 6x^2y\frac{dy}{dx} = 6x^4 + y\cos\left(\frac{y}{x}\right)xcos(xy​)dxdy​+6x2ydxdy​=6x4+ycos(xy​)
  5. Factor out dydx\frac{dy}{dx}dxdy​ and solve:
    dydx[xcos⁡(yx)+6x2y]=6x4+ycos⁡(yx)\frac{dy}{dx}\left[ x\cos\left(\frac{y}{x}\right) + 6x^2y \right] = 6x^4 + y\cos\left(\frac{y}{x}\right)dxdy​[xcos(xy​)+6x2y]=6x4+ycos(xy​)
    dydx=6x4+ycos⁡(yx)xcos⁡(yx)+6x2y\frac{dy}{dx} = \frac{6x^4 + y\cos\left(\frac{y}{x}\right)}{x\cos\left(\frac{y}{x}\right) + 6x^2y}dxdy​=xcos(xy​)+6x2y6x4+ycos(xy​)​
Show your steps and calculations here.

Practice Problem 5: Tangent Line to an Implicit Transcendental Curve

Consider the curve defined by the equation y+sin⁡y=x+πy + \sin y = x + \piy+siny=x+π. Find the equation of the tangent line to the curve at the point (π,2π)(\pi, 2\pi)(π,2π) in both point-slope and slope-intercept forms.

Solution Key:
  1. Governing Formula: The Point-Slope Equation of a Tangent Line:
    y−y1=m(x−x1),m=dydx∣(x1,y1)y - y_1 = m(x - x_1), \qquad m = \left.\frac{dy}{dx}\right|_{(x_1, y_1)}y−y1​=m(x−x1​),m=dxdy​​(x1​,y1​)​
    (This formula is defined in Implicit Differentiation.)
  2. Differentiate the curve equation with respect to xxx:
    ddx[y]+ddx[sin⁡y]=ddx[x]+ddx[π]\frac{d}{dx}[y] + \frac{d}{dx}[\sin y] = \frac{d}{dx}[x] + \frac{d}{dx}[\pi]dxd​[y]+dxd​[siny]=dxd​[x]+dxd​[π]
    dydx+cos⁡(y)dydx=1+0\frac{dy}{dx} + \cos(y)\frac{dy}{dx} = 1 + 0dxdy​+cos(y)dxdy​=1+0
  3. Isolate dydx\frac{dy}{dx}dxdy​:
    dydx(1+cos⁡y)=1  ⟹  dydx=11+cos⁡y\frac{dy}{dx}(1 + \cos y) = 1 \implies \frac{dy}{dx} = \frac{1}{1 + \cos y}dxdy​(1+cosy)=1⟹dxdy​=1+cosy1​
  4. Evaluate slope at (π,2π)(\pi, 2\pi)(π,2π):
    m=dydx∣(π,2π)=11+cos⁡(2π)=11+1=12m = \left.\frac{dy}{dx}\right|_{(\pi, 2\pi)} = \frac{1}{1 + \cos(2\pi)} = \frac{1}{1 + 1} = \frac{1}{2}m=dxdy​​(π,2π)​=1+cos(2π)1​=1+11​=21​
  5. Assemble Tangent Line Equations:
    • Point-Slope Form:
      y−2π=12(x−π)y - 2\pi = \frac{1}{2}(x - \pi)y−2π=21​(x−π)
    • Slope-Intercept Form:
      y=12x−π2+2π  ⟹  y=12x+3π2y = \frac{1}{2}x - \frac{\pi}{2} + 2\pi \implies y = \frac{1}{2}x + \frac{3\pi}{2}y=21​x−2π​+2π⟹y=21​x+23π​
Show your steps and calculations here.

Regroup 1

  • Review student solutions on collecting dydx\frac{dy}{dx}dxdy​ terms. Emphasize that every term containing dydx\frac{dy}{dx}dxdy​ must be moved to one side before factoring.
  • Clarify common sign errors when applying the Product Rule on expressions like −xy-xy−xy (distributing the negative sign: ddx[−xy]=−y−xdydx\frac{d}{dx}[-xy] = -y - x\frac{dy}{dx}dxd​[−xy]=−y−xdxdy​).

Skill Block 2

Worked Example A: Handling Variable Base and Variable Exponent Functions

Find dydx\frac{dy}{dx}dxdy​ for the function y=xxy = x^xy=xx (for x>0x > 0x>0).

Worked Solution:
  1. Governing Formula: The Logarithmic Differentiation Process:
    ln⁡y=ln⁡[f(x)],ddx[ln⁡y]=1ydydx\ln y = \ln[f(x)], \qquad \frac{d}{dx}[\ln y] = \frac{1}{y}\frac{dy}{dx}lny=ln[f(x)],dxd​[lny]=y1​dxdy​
    (This technique is defined in Logarithmic Differentiation.)
  2. Apply the natural logarithm to both sides:
    ln⁡y=ln⁡(xx)\ln y = \ln(x^x)lny=ln(xx)
  3. Bring down the exponent using the Power Property:
    ln⁡y=x⋅ln⁡x\ln y = x \cdot \ln xlny=x⋅lnx
  4. Differentiate implicitly with respect to xxx: Apply the Product Rule on the right side:
    1ydydx=ddx[x]⋅ln⁡x+x⋅ddx[ln⁡x]=1⋅ln⁡x+x⋅(1x)=ln⁡x+1\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[x]\cdot \ln x + x\cdot \frac{d}{dx}[\ln x] = 1 \cdot \ln x + x\cdot\left(\frac{1}{x}\right) = \ln x + 1y1​dxdy​=dxd​[x]⋅lnx+x⋅dxd​[lnx]=1⋅lnx+x⋅(x1​)=lnx+1
  5. Multiply by yyy and back-substitute y=xxy = x^xy=xx:
    dydx=y(ln⁡x+1)=xx(ln⁡x+1)\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)dxdy​=y(lnx+1)=xx(lnx+1)

Worked Example B: Simplifying Multi-Factor Radical and Quotient Expressions

Find dydx\frac{dy}{dx}dxdy​ for the function y=x2x−73(x3+1)4y = \frac{x \sqrt[3]{2x-7}}{(x^3 + 1)^4}y=(x3+1)4x32x−7​​.

Worked Solution:
  1. Governing Formula: Logarithmic Expansion Properties:
    ln⁡[A⋅B1/3C4]=ln⁡A+13ln⁡B−4ln⁡C\ln\left[\frac{A \cdot B^{1/3}}{C^4}\right] = \ln A + \frac{1}{3}\ln B - 4\ln Cln[C4A⋅B1/3​]=lnA+31​lnB−4lnC
    (These properties are defined in Logarithmic and Exponential Properties and Logarithmic Differentiation.)
  2. Take the natural logarithm of both sides:
    ln⁡y=ln⁡[x(2x−7)1/3(x3+1)4]\ln y = \ln\left[ \frac{x (2x-7)^{1/3}}{(x^3+1)^4} \right]lny=ln[(x3+1)4x(2x−7)1/3​]
  3. Expand using log properties:
    ln⁡y=ln⁡x+13ln⁡(2x−7)−4ln⁡(x3+1)\ln y = \ln x + \frac{1}{3}\ln(2x-7) - 4\ln(x^3+1)lny=lnx+31​ln(2x−7)−4ln(x3+1)
  4. Differentiate implicitly with respect to xxx:
    1ydydx=1x+13(22x−7)−4(3x2x3+1)=1x+23(2x−7)−12x2x3+1\frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + \frac{1}{3}\left(\frac{2}{2x-7}\right) - 4\left(\frac{3x^2}{x^3+1}\right) = \frac{1}{x} + \frac{2}{3(2x-7)} - \frac{12x^2}{x^3+1}y1​dxdy​=x1​+31​(2x−72​)−4(x3+13x2​)=x1​+3(2x−7)2​−x3+112x2​
  5. Multiply by yyy and back-substitute:
    dydx=x2x−73(x3+1)4[1x+23(2x−7)−12x2x3+1]\frac{dy}{dx} = \frac{x \sqrt[3]{2x-7}}{(x^3+1)^4} \left[ \frac{1}{x} + \frac{2}{3(2x-7)} - \frac{12x^2}{x^3+1} \right]dxdy​=(x3+1)4x32x−7​​[x1​+3(2x−7)2​−x3+112x2​]

Practice Problem 6: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

  • (a) In the Conceptual Framework, why does treating y=xxy = x^xy=xx with the Power Rule (dydx=x⋅xx−1=xx\frac{dy}{dx} = x \cdot x^{x-1} = x^xdxdy​=x⋅xx−1=xx) or Exponential Rule (dydx=ln⁡(x)xx\frac{dy}{dx} = \ln(x)x^xdxdy​=ln(x)xx) produce incorrect results?
  • (b) Deconstruct Worked Example B: Contrast the logarithmic approach with computing the derivative directly via the Quotient and Product rules. How does logarithmic pre-processing simplify the algebra?
  • (c) Review the linked Engineering Applications (Materials Engineering (Non-Linear Fractional Material Strain)). Why does the materials engineer evaluate the fractional rate of change 1vdvdT\frac{1}{v}\frac{dv}{dT}v1​dTdv​ directly without multiplying by v(T)v(T)v(T)?
Solution Key:
  • (a) In elementary calculus, the Power Rule ddx[xn]=nxn−1\frac{d}{dx}[x^n] = n x^{n-1}dxd​[xn]=nxn−1 is proven strictly under the condition that the power nnn is a constant number. The Exponential Rule ddx[bx]=ln⁡(b)bx\frac{d}{dx}[b^x] = \ln(b)b^xdxd​[bx]=ln(b)bx requires that the base bbb is a constant. In y=xxy = x^xy=xx, both the base and the exponent change simultaneously as xxx varies. The correct derivative xx(ln⁡x+1)=xxln⁡x+xxx^x(\ln x + 1) = x^x\ln x + x^xxx(lnx+1)=xxlnx+xx is actually the sum of treating the base as variable plus treating the exponent as variable—a result that can only be proven via logarithmic differentiation.
  • (b) In Worked Example B, applying the Quotient Rule directly would require setting up u′v−uv′v2\frac{u'v - uv'}{v^2}v2u′v−uv′​ where u=x(2x−7)1/3u = x(2x-7)^{1/3}u=x(2x−7)1/3 requires the Product Rule and v=(x3+1)4v = (x^3+1)^4v=(x3+1)4 requires the Chain Rule, producing massive fractions with compound common denominators. By taking the logarithm first, the expression expands into ln⁡x+13ln⁡(2x−7)−4ln⁡(x3+1)\ln x + \frac{1}{3}\ln(2x-7) - 4\ln(x^3+1)lnx+31​ln(2x−7)−4ln(x3+1). Each term is then differentiated in a single line without any common denominator algebra.
  • (c) In Materials Engineering (Non-Linear Fractional Material Strain), the volume function is v(T)=(T2+1)ln⁡Tv(T) = (T^2+1)^{\ln T}v(T)=(T2+1)lnT. The engineer is tasked with finding the fractional rate of strain 1vdvdT\frac{1}{v}\frac{dv}{dT}v1​dTdv​, which represents the localized percentage expansion per degree Kelvin. Because differentiating ln⁡v\ln vlnv naturally yields 1vdvdT\frac{1}{v}\frac{dv}{dT}v1​dTdv​ on the left side, the fractional rate is obtained immediately without needing to multiply back by the volume function v(T)v(T)v(T).
Show your steps and calculations here.

Practice Problem 7: Logarithmic Differentiation of Variable Exponent Function

Find the derivative dydx\frac{dy}{dx}dxdy​ of the function:

y=xln⁡x(x>0)y = x^{\ln x} \qquad (x > 0)y=xlnx(x>0)

Solution Key:
  1. Governing Formula: The Power Property of Logarithms and Logarithmic Differentiation:
    ln⁡(AB)=Bln⁡A,ddx[ln⁡y]=1ydydx\ln(A^B) = B\ln A, \qquad \frac{d}{dx}[\ln y] = \frac{1}{y}\frac{dy}{dx}ln(AB)=BlnA,dxd​[lny]=y1​dxdy​
    (These rules are defined in Logarithmic Differentiation.)
  2. Apply the natural logarithm to both sides:
    ln⁡y=ln⁡(xln⁡x)\ln y = \ln\left(x^{\ln x}\right)lny=ln(xlnx)
  3. Bring down the exponent using the Power Property:
    ln⁡y=(ln⁡x)⋅(ln⁡x)=(ln⁡x)2\ln y = (\ln x) \cdot (\ln x) = (\ln x)^2lny=(lnx)⋅(lnx)=(lnx)2
  4. Differentiate implicitly with respect to xxx using the Chain Rule:
    1ydydx=2(ln⁡x)⋅ddx[ln⁡x]=2(ln⁡x)⋅1x=2ln⁡xx\frac{1}{y}\frac{dy}{dx} = 2(\ln x) \cdot \frac{d}{dx}[\ln x] = 2(\ln x) \cdot \frac{1}{x} = \frac{2\ln x}{x}y1​dxdy​=2(lnx)⋅dxd​[lnx]=2(lnx)⋅x1​=x2lnx​
  5. Multiply by yyy and back-substitute y=xln⁡xy = x^{\ln x}y=xlnx:
    dydx=y⋅(2ln⁡xx)=xln⁡x⋅2ln⁡xx\frac{dy}{dx} = y \cdot \left(\frac{2\ln x}{x}\right) = x^{\ln x} \cdot \frac{2\ln x}{x}dxdy​=y⋅(x2lnx​)=xlnx⋅x2lnx​
Show your steps and calculations here.

Practice Problem 8: Logarithmic Differentiation of Trigonometric Base and Exponent

Find the derivative dydx\frac{dy}{dx}dxdy​ of the function:

y=(cos⁡x)sin⁡x(0<x<π2)y = (\cos x)^{\sin x} \qquad \left(0 < x < \frac{\pi}{2}\right)y=(cosx)sinx(0<x<2π​)

Solution Key:
  1. Governing Formula: Logarithmic Differentiation with Product Rule:
    ln⁡y=g(x)ln⁡[f(x)]  ⟹  1ydydx=g′(x)ln⁡[f(x)]+g(x)f′(x)f(x)\ln y = g(x)\ln[f(x)] \implies \frac{1}{y}\frac{dy}{dx} = g'(x)\ln[f(x)] + g(x)\frac{f'(x)}{f(x)}lny=g(x)ln[f(x)]⟹y1​dxdy​=g′(x)ln[f(x)]+g(x)f(x)f′(x)​
    (This formula is defined in Logarithmic Differentiation.)
  2. Apply the natural logarithm to both sides:
    ln⁡y=ln⁡[(cos⁡x)sin⁡x]=sin⁡(x)⋅ln⁡(cos⁡x)\ln y = \ln\left[(\cos x)^{\sin x}\right] = \sin(x) \cdot \ln(\cos x)lny=ln[(cosx)sinx]=sin(x)⋅ln(cosx)
  3. Differentiate implicitly with respect to xxx using the Product Rule:
    1ydydx=ddx[sin⁡x]⋅ln⁡(cos⁡x)+sin⁡(x)⋅ddx[ln⁡(cos⁡x)]\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[\sin x]\cdot \ln(\cos x) + \sin(x)\cdot \frac{d}{dx}[\ln(\cos x)]y1​dxdy​=dxd​[sinx]⋅ln(cosx)+sin(x)⋅dxd​[ln(cosx)]
    Recall that ddx[ln⁡(cos⁡x)]=−sin⁡xcos⁡x\frac{d}{dx}[\ln(\cos x)] = \frac{-\sin x}{\cos x}dxd​[ln(cosx)]=cosx−sinx​:
    1ydydx=cos⁡(x)ln⁡(cos⁡x)+sin⁡(x)(−sin⁡xcos⁡x)=cos⁡(x)ln⁡(cos⁡x)−sin⁡2xcos⁡x\frac{1}{y}\frac{dy}{dx} = \cos(x)\ln(\cos x) + \sin(x)\left(\frac{-\sin x}{\cos x}\right) = \cos(x)\ln(\cos x) - \frac{\sin^2 x}{\cos x}y1​dxdy​=cos(x)ln(cosx)+sin(x)(cosx−sinx​)=cos(x)ln(cosx)−cosxsin2x​
  4. Multiply by yyy and back-substitute y=(cos⁡x)sin⁡xy = (\cos x)^{\sin x}y=(cosx)sinx:
    dydx=(cos⁡x)sin⁡x[cos⁡(x)ln⁡(cos⁡x)−sin⁡2xcos⁡x]\frac{dy}{dx} = (\cos x)^{\sin x} \left[ \cos(x)\ln(\cos x) - \frac{\sin^2 x}{\cos x} \right]dxdy​=(cosx)sinx[cos(x)ln(cosx)−cosxsin2x​]
Show your steps and calculations here.

Practice Problem 9: Logarithmic Differentiation of Rational Base with Power Exponent

Find the derivative dydx\frac{dy}{dx}dxdy​ of the function:

y=(x+1x2+1)x3y = \left(\frac{x+1}{x^2+1}\right)^{x^3}y=(x2+1x+1​)x3

Solution Key:
  1. Governing Formula: Logarithmic Differentiation with Power and Quotient Properties:
    ln⁡[(AB)C]=C[ln⁡A−ln⁡B]\ln\left[\left(\frac{A}{B}\right)^C\right] = C[\ln A - \ln B]ln[(BA​)C]=C[lnA−lnB]
    (These properties are defined in Logarithmic Differentiation.)
  2. Apply the natural logarithm and expand:
    ln⁡y=ln⁡[(x+1x2+1)x3]=x3⋅[ln⁡(x+1)−ln⁡(x2+1)]\ln y = \ln\left[ \left(\frac{x+1}{x^2+1}\right)^{x^3} \right] = x^3 \cdot \left[ \ln(x+1) - \ln(x^2+1) \right]lny=ln[(x2+1x+1​)x3]=x3⋅[ln(x+1)−ln(x2+1)]
  3. Differentiate implicitly with respect to xxx using the Product Rule:
    1ydydx=ddx[x3]⋅[ln⁡(x+1)−ln⁡(x2+1)]+x3⋅ddx[ln⁡(x+1)−ln⁡(x2+1)]\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[x^3]\cdot \left[\ln(x+1) - \ln(x^2+1)\right] + x^3 \cdot \frac{d}{dx}\left[\ln(x+1) - \ln(x^2+1)\right]y1​dxdy​=dxd​[x3]⋅[ln(x+1)−ln(x2+1)]+x3⋅dxd​[ln(x+1)−ln(x2+1)]
    1ydydx=3x2[ln⁡(x+1)−ln⁡(x2+1)]+x3[1x+1−2xx2+1]\frac{1}{y}\frac{dy}{dx} = 3x^2 \left[\ln(x+1) - \ln(x^2+1)\right] + x^3 \left[ \frac{1}{x+1} - \frac{2x}{x^2+1} \right]y1​dxdy​=3x2[ln(x+1)−ln(x2+1)]+x3[x+11​−x2+12x​]
  4. Recombine the logarithm and multiply by yyy:
    dydx=(x+1x2+1)x3{3x2ln⁡(x+1x2+1)+x3[1x+1−2xx2+1]}\frac{dy}{dx} = \left(\frac{x+1}{x^2+1}\right)^{x^3} \left\{ 3x^2 \ln\left(\frac{x+1}{x^2+1}\right) + x^3 \left[ \frac{1}{x+1} - \frac{2x}{x^2+1} \right] \right\}dxdy​=(x2+1x+1​)x3{3x2ln(x2+1x+1​)+x3[x+11​−x2+12x​]}
Show your steps and calculations here.

Regroup 2

  • Review logarithmic expansion rules before differentiation. Ensure students do not attempt to differentiate [f(x)]g(x)[f(x)]^{g(x)}[f(x)]g(x) by pulling g(x)g(x)g(x) in front and subtracting 1 from the exponent (a common misapplication of the Power Rule).
  • Emphasize the back-substitution step: the final answer for dydx\frac{dy}{dx}dxdy​ in logarithmic differentiation must have yyy replaced by the original function expression f(x)f(x)f(x) so it remains purely in terms of xxx.

Skill Block 3

Worked Example A: Second Derivative of a Circle with Constraint Simplification

Find d2ydx2\frac{d^2y}{dx^2}dx2d2y​ for the circle relation x2+y2=1x^2 + y^2 = 1x2+y2=1.

Worked Solution:
  1. Governing Formula: The Quotient Rule for Higher-Order Derivatives:
    d2ydx2=ddx[dydx]=u′v−uv′v2\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{u'v - uv'}{v^2}dx2d2y​=dxd​[dxdy​]=v2u′v−uv′​
    (This formula is defined in Higher-Order Implicit Differentiation.)
  2. Find the first derivative dydx\frac{dy}{dx}dxdy​:
    2x+2ydydx=0  ⟹  2ydydx=−2x  ⟹  dydx=−xy2x + 2y\frac{dy}{dx} = 0 \implies 2y\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}2x+2ydxdy​=0⟹2ydxdy​=−2x⟹dxdy​=−yx​
  3. Differentiate a second time using the Quotient Rule:
    d2ydx2=ddx[−xy]=−[(1)(y)−(x)(dydx)y2]=−y+xdydxy2\frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{x}{y}\right] = -\left[ \frac{(1)(y) - (x)\left(\frac{dy}{dx}\right)}{y^2} \right] = \frac{-y + x\frac{dy}{dx}}{y^2}dx2d2y​=dxd​[−yx​]=−​y2(1)(y)−(x)(dxdy​)​​=y2−y+xdxdy​​
  4. Back-substitute dydx=−xy\frac{dy}{dx} = -\frac{x}{y}dxdy​=−yx​:
    d2ydx2=−y+x(−xy)y2=−y−x2yy2\frac{d^2y}{dx^2} = \frac{-y + x\left(-\frac{x}{y}\right)}{y^2} = \frac{-y - \frac{x^2}{y}}{y^2}dx2d2y​=y2−y+x(−yx​)​=y2−y−yx2​​
  5. Clear fractions and substitute the original constraint (x2+y2=1x^2 + y^2 = 1x2+y2=1): Multiply numerator and denominator by yyy:
    d2ydx2=−y2−x2y3=−(x2+y2)y3\frac{d^2y}{dx^2} = \frac{-y^2 - x^2}{y^3} = \frac{-(x^2 + y^2)}{y^3}dx2d2y​=y3−y2−x2​=y3−(x2+y2)​
    Substitute x2+y2=1x^2 + y^2 = 1x2+y2=1:
    d2ydx2=−(1)y3=−1y3\frac{d^2y}{dx^2} = \frac{-(1)}{y^3} = -\frac{1}{y^3}dx2d2y​=y3−(1)​=−y31​

Worked Example B: Second Derivative of Higher-Degree Polynomial Relations

Find d2ydx2\frac{d^2y}{dx^2}dx2d2y​ for the implicit relation 2x3−3y2=82x^3 - 3y^2 = 82x3−3y2=8.

Worked Solution:
  1. Governing Formula: The Quotient Rule with Implicit Chain Substitution:
    d2ydx2=ddx[x2y]\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{x^2}{y}\right]dx2d2y​=dxd​[yx2​]
    (This rule is defined in Higher-Order Implicit Differentiation.)
  2. Find the first derivative dydx\frac{dy}{dx}dxdy​:
    6x2−6ydydx=0  ⟹  −6ydydx=−6x2  ⟹  dydx=x2y6x^2 - 6y\frac{dy}{dx} = 0 \implies -6y\frac{dy}{dx} = -6x^2 \implies \frac{dy}{dx} = \frac{x^2}{y}6x2−6ydxdy​=0⟹−6ydxdy​=−6x2⟹dxdy​=yx2​
  3. Differentiate a second time using the Quotient Rule:
    d2ydx2=ddx[x2]⋅y−x2⋅ddx[y]y2=2xy−x2(dydx)y2\frac{d^2y}{dx^2} = \frac{\frac{d}{dx}[x^2]\cdot y - x^2\cdot \frac{d}{dx}[y]}{y^2} = \frac{2xy - x^2\left(\frac{dy}{dx}\right)}{y^2}dx2d2y​=y2dxd​[x2]⋅y−x2⋅dxd​[y]​=y22xy−x2(dxdy​)​
  4. Back-substitute dydx=x2y\frac{dy}{dx} = \frac{x^2}{y}dxdy​=yx2​:
    d2ydx2=2xy−x2(x2y)y2=2xy−x4yy2\frac{d^2y}{dx^2} = \frac{2xy - x^2\left(\frac{x^2}{y}\right)}{y^2} = \frac{2xy - \frac{x^4}{y}}{y^2}dx2d2y​=y22xy−x2(yx2​)​=y22xy−yx4​​
  5. Clear the complex fraction by multiplying top and bottom by yyy:
    d2ydx2=2xy2−x4y3\frac{d^2y}{dx^2} = \frac{2xy^2 - x^4}{y^3}dx2d2y​=y32xy2−x4​

Practice Problem 10: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

  • (a) In the Conceptual Framework, why is leaving a dydx\frac{dy}{dx}dxdy​ term inside your final answer for d2ydx2\frac{d^2y}{dx^2}dx2d2y​ considered incomplete?
  • (b) Deconstruct Worked Example A: In Step 5, how does recognizing the circle's original equation x2+y2=1x^2 + y^2 = 1x2+y2=1 simplify the complex algebraic fraction −y2−x2y3\frac{-y^2 - x^2}{y^3}y3−y2−x2​ into −1y3-\frac{1}{y^3}−y31​?
  • (c) Review the linked Engineering Applications (Aerospace Engineering (Structural Integrity under Thermal Expansion)). What physical property of the titanium bulkhead viewport does the second derivative d2ydx2=−2564\frac{d^2y}{dx^2} = -\frac{25}{64}dx2d2y​=−6425​ quantify during atmospheric re-entry?
Solution Key:
  • (a) The purpose of computing d2ydx2\frac{d^2y}{dx^2}dx2d2y​ is to establish a closed-form formula that evaluates the second derivative directly from coordinate data (x,y)(x, y)(x,y). Because dydx\frac{dy}{dx}dxdy​ is an intermediate rate, leaving it un-substituted leaves the formula unresolved. Performing back-substitution ensures that the second derivative is a self-contained function of the curve coordinates.
  • (b) In Worked Example A, clearing the complex fraction produced −y2−x2y3\frac{-y^2 - x^2}{y^3}y3−y2−x2​. Factoring out −1-1−1 gives −(x2+y2)y3\frac{-(x^2 + y^2)}{y^3}y3−(x2+y2)​. Because the point (x,y)(x, y)(x,y) is constrained to lie on the unit circle x2+y2=1x^2 + y^2 = 1x2+y2=1, the quantity (x2+y2)(x^2 + y^2)(x2+y2) in the numerator can be replaced with the constant 111, yielding the elegant compressed result −1y3-\frac{1}{y^3}−y31​.
  • (c) In Aerospace Engineering (Structural Integrity under Thermal Expansion), aerodynamic heating during atmospheric re-entry causes the titanium bulkhead window frame to warp along x2+y2=25x^2 + y^2 = 25x2+y2=25. The negative second derivative d2ydx2=−2564≈−0.3906 cm−1\frac{d^2y}{dx^2} = -\frac{25}{64} \approx -0.3906\text{ cm}^{-1}dx2d2y​=−6425​≈−0.3906 cm−1 indicates downward concavity and directly determines the physical curvature κ=∣y′′∣(1+(y′)2)3/2\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}κ=(1+(y′)2)3/2∣y′′∣​. Curvature spikes amplify localized shear stresses; calculating this value confirms the window seal will not rupture under re-entry pressure loads.
Show your steps and calculations here.

Practice Problem 11: Second Derivative of a Hyperbola with Geometric Simplification

For the hyperbola implicitly defined by the equation x2−y2=1x^2 - y^2 = 1x2−y2=1, find the second derivative d2ydx2\frac{d^2y}{dx^2}dx2d2y​ and simplify your answer using the original equation constraint.

Solution Key:
  1. Governing Formula: The Quotient Rule for Higher-Order Derivatives:
    d2ydx2=ddx[dydx]=u′v−uv′v2\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{u'v - uv'}{v^2}dx2d2y​=dxd​[dxdy​]=v2u′v−uv′​
    (This formula is defined in Higher-Order Implicit Differentiation.)
  2. Find the first derivative dydx\frac{dy}{dx}dxdy​:
    ddx[x2]−ddx[y2]=ddx[1]\frac{d}{dx}[x^2] - \frac{d}{dx}[y^2] = \frac{d}{dx}[1]dxd​[x2]−dxd​[y2]=dxd​[1]
    2x−2ydydx=0  ⟹  2ydydx=2x  ⟹  dydx=xy2x - 2y\frac{dy}{dx} = 0 \implies 2y\frac{dy}{dx} = 2x \implies \frac{dy}{dx} = \frac{x}{y}2x−2ydxdy​=0⟹2ydxdy​=2x⟹dxdy​=yx​
  3. Differentiate dydx\frac{dy}{dx}dxdy​ using the Quotient Rule:
    d2ydx2=ddx[x]⋅y−x⋅ddx[y]y2=1⋅y−x(dydx)y2\frac{d^2y}{dx^2} = \frac{\frac{d}{dx}[x]\cdot y - x\cdot \frac{d}{dx}[y]}{y^2} = \frac{1\cdot y - x\left(\frac{dy}{dx}\right)}{y^2}dx2d2y​=y2dxd​[x]⋅y−x⋅dxd​[y]​=y21⋅y−x(dxdy​)​
  4. Back-substitute dydx=xy\frac{dy}{dx} = \frac{x}{y}dxdy​=yx​:
    d2ydx2=y−x(xy)y2=y−x2yy2\frac{d^2y}{dx^2} = \frac{y - x\left(\frac{x}{y}\right)}{y^2} = \frac{y - \frac{x^2}{y}}{y^2}dx2d2y​=y2y−x(yx​)​=y2y−yx2​​
  5. Clear fractions and substitute the hyperbola constraint (x2−y2=1x^2 - y^2 = 1x2−y2=1): Multiply numerator and denominator by yyy:
    d2ydx2=y2−x2y3\frac{d^2y}{dx^2} = \frac{y^2 - x^2}{y^3}dx2d2y​=y3y2−x2​
    Since x2−y2=1x^2 - y^2 = 1x2−y2=1, multiplying by −1-1−1 gives y2−x2=−1y^2 - x^2 = -1y2−x2=−1:
    d2ydx2=−1y3=−1y3\frac{d^2y}{dx^2} = \frac{-1}{y^3} = -\frac{1}{y^3}dx2d2y​=y3−1​=−y31​
Show your steps and calculations here.

Practice Problem 12: Second Derivative of a Transcendental Relation via Chain Rule

For the curve implicitly defined by y+cos⁡y=x−π3y + \cos y = x - \frac{\pi}{3}y+cosy=x−3π​, find the second derivative d2ydx2\frac{d^2y}{dx^2}dx2d2y​ in terms of yyy.

Solution Key:
  1. Governing Formula: The Power Chain Rule for Higher-Order Derivatives:
    ddx[(g(y))−1]=−1(g(y))−2⋅g′(y)dydx\frac{d}{dx}[(g(y))^{-1}] = -1(g(y))^{-2} \cdot g'(y)\frac{dy}{dx}dxd​[(g(y))−1]=−1(g(y))−2⋅g′(y)dxdy​
    (This rule is defined in Higher-Order Implicit Differentiation.)
  2. Find the first derivative dydx\frac{dy}{dx}dxdy​:
    ddx[y]+ddx[cos⁡y]=ddx[x]−ddx[π3]\frac{d}{dx}[y] + \frac{d}{dx}[\cos y] = \frac{d}{dx}[x] - \frac{d}{dx}\left[\frac{\pi}{3}\right]dxd​[y]+dxd​[cosy]=dxd​[x]−dxd​[3π​]
    dydx−sin⁡(y)dydx=1−0\frac{dy}{dx} - \sin(y)\frac{dy}{dx} = 1 - 0dxdy​−sin(y)dxdy​=1−0
    dydx(1−sin⁡y)=1  ⟹  dydx=11−sin⁡y=(1−sin⁡y)−1\frac{dy}{dx}(1 - \sin y) = 1 \implies \frac{dy}{dx} = \frac{1}{1 - \sin y} = (1 - \sin y)^{-1}dxdy​(1−siny)=1⟹dxdy​=1−siny1​=(1−siny)−1
  3. Differentiate a second time with respect to xxx using the Chain Rule:
    d2ydx2=ddx[(1−sin⁡y)−1]=−1(1−sin⁡y)−2⋅ddx[1−sin⁡y]\frac{d^2y}{dx^2} = \frac{d}{dx}\left[(1 - \sin y)^{-1}\right] = -1(1 - \sin y)^{-2} \cdot \frac{d}{dx}[1 - \sin y]dx2d2y​=dxd​[(1−siny)−1]=−1(1−siny)−2⋅dxd​[1−siny]
    d2ydx2=−1(1−sin⁡y)2⋅(−cos⁡(y)dydx)=cos⁡y(1−sin⁡y)2⋅dydx\frac{d^2y}{dx^2} = -\frac{1}{(1 - \sin y)^2} \cdot \left(-\cos(y)\frac{dy}{dx}\right) = \frac{\cos y}{(1 - \sin y)^2} \cdot \frac{dy}{dx}dx2d2y​=−(1−siny)21​⋅(−cos(y)dxdy​)=(1−siny)2cosy​⋅dxdy​
  4. Back-substitute dydx=11−sin⁡y\frac{dy}{dx} = \frac{1}{1 - \sin y}dxdy​=1−siny1​:
    d2ydx2=cos⁡y(1−sin⁡y)2⋅(11−sin⁡y)=cos⁡y(1−sin⁡y)3\frac{d^2y}{dx^2} = \frac{\cos y}{(1 - \sin y)^2} \cdot \left(\frac{1}{1 - \sin y}\right) = \frac{\cos y}{(1 - \sin y)^3}dx2d2y​=(1−siny)2cosy​⋅(1−siny1​)=(1−siny)3cosy​
Show your steps and calculations here.

Regroup 3

  • Emphasize the algebraic payoff of simplifying higher-order derivatives with original curve equations (x2+y2=r2x^2 + y^2 = r^2x2+y2=r2, x2−y2=1x^2 - y^2 = 1x2−y2=1). Show how this geometric substitution turns unwieldy fractions into compact terms.
  • Review student work on Problem 12: demonstrate why writing dydx=(1−sin⁡y)−1\frac{dy}{dx} = (1 - \sin y)^{-1}dxdy​=(1−siny)−1 and using the Power Chain Rule is often faster and less error-prone than applying the Quotient Rule.

Synthesis Wrap-up

  • Core Takeaways from Session 09:
    1. Implicit Differentiation: Treat yyy as y(x)y(x)y(x); append dydx\frac{dy}{dx}dxdy​ whenever differentiating yyy. Collect, factor, and solve.
    2. Logarithmic Differentiation: The indispensable tool for variable towers (f(x)g(x)f(x)^{g(x)}f(x)g(x)) and multi-factor rational systems. Expand with log properties before differentiating.
    3. Higher-Order Derivatives: Differentiate dydx\frac{dy}{dx}dxdy​ via quotient or power chain rules, back-substitute dydx\frac{dy}{dx}dxdy​, and simplify with original curve constraints.
  • Homework Preparation: Complete Knewton Alta assignments on Implicit Differentiation and Logarithmic Differentiation. Review tangent line setups and practice checking that all final derivatives for logarithmic differentiation are expressed purely in terms of xxx.