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Derivatives of Logarithmic Functions

Formulas & Definitions
Core Definition

For any real argument x>0x > 0x>0: Natural Logarithm:

ddx[ln⁡(x)]=1x\frac{d}{dx}[\ln(x)] = \frac{1}{x}dxd​[ln(x)]=x1​
General Logarithm (Base a>0,a≠1a > 0, a \neq 1a>0,a=1):
ddx[log⁡a(x)]=1xln⁡(a)\frac{d}{dx}[\log_a(x)] = \frac{1}{x \ln(a)}dxd​[loga​(x)]=xln(a)1​
When layered with a differentiable inner function u=g(x)>0u = g(x) > 0u=g(x)>0, the Chain Rule yields:
ddx[ln⁡(u)]=1ududx=u′uandddx[log⁡a(u)]=u′uln⁡(a)\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx} = \frac{u'}{u} \qquad \text{and} \qquad \frac{d}{dx}[\log_a(u)] = \frac{u'}{u \ln(a)}dxd​[ln(u)]=u1​dxdu​=uu′​anddxd​[loga​(u)]=uln(a)u′​

Concept Explanation

1. Derivation via Implicit Differentiation

The derivative of ln⁡(x)\ln(x)ln(x) is established directly from the fact that y=ln⁡(x)y = \ln(x)y=ln(x) is the inverse of x=eyx = e^yx=ey. Differentiating both sides implicitly with respect to xxx:

ddx[ey]=ddx[x]  ⟹  eydydx=1  ⟹  dydx=1ey\frac{d}{dx}[e^y] = \frac{d}{dx}[x] \implies e^y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{e^y}dxd​[ey]=dxd​[x]⟹eydxdy​=1⟹dxdy​=ey1​
Because ey=xe^y = xey=x, substituting back gives dydx=1x\frac{dy}{dx} = \frac{1}{x}dxdy​=x1​.

2. General Logarithm via Change of Base

By the Change of Base formula (see Logarithmic and Exponential Properties), log⁡a(x)=ln⁡(x)ln⁡(a)\log_a(x) = \frac{\ln(x)}{\ln(a)}loga​(x)=ln(a)ln(x)​. Differentiating using the Constant Multiple Rule:

ddx[log⁡a(x)]=1ln⁡(a)⋅ddx[ln⁡(x)]=1ln⁡(a)⋅1x=1xln⁡(a)\frac{d}{dx}[\log_a(x)] = \frac{1}{\ln(a)} \cdot \frac{d}{dx}[\ln(x)] = \frac{1}{\ln(a)} \cdot \frac{1}{x} = \frac{1}{x \ln(a)}dxd​[loga​(x)]=ln(a)1​⋅dxd​[ln(x)]=ln(a)1​⋅x1​=xln(a)1​

3. Logarithmic Pre-Processing: The Time-Saving Technique

Before applying calculus derivative rules, always expand the argument completely using fundamental logarithm properties (see Logarithmic and Exponential Properties and Negative and Rational Exponents):

  • Product Property: ln⁡(A⋅B)=ln⁡(A)+ln⁡(B)  ⟹  ddx[ln⁡(AB)]=A′A+B′B\ln(A \cdot B) = \ln(A) + \ln(B) \implies \frac{d}{dx}[\ln(AB)] = \frac{A'}{A} + \frac{B'}{B}ln(A⋅B)=ln(A)+ln(B)⟹dxd​[ln(AB)]=AA′​+BB′​
  • Quotient Property: ln⁡(AB)=ln⁡(A)−ln⁡(B)  ⟹  ddx[ln⁡(AB)]=A′A−B′B\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B) \implies \frac{d}{dx}\left[\ln\left(\frac{A}{B}\right)\right] = \frac{A'}{A} - \frac{B'}{B}ln(BA​)=ln(A)−ln(B)⟹dxd​[ln(BA​)]=AA′​−BB′​
  • Power Property: ln⁡(Ak)=kln⁡(A)  ⟹  ddx[ln⁡(Ak)]=kA′A\ln(A^k) = k \ln(A) \implies \frac{d}{dx}[\ln(A^k)] = k \frac{A'}{A}ln(Ak)=kln(A)⟹dxd​[ln(Ak)]=kAA′​

Differentiating the expanded sum of terms eliminates the need to execute cumbersome Quotient and Product rules inside u′u\frac{u'}{u}uu′​.


Engineering Applications

  • Aerospace Engineering (Tsiolkovsky Rocket Equation): The rocket velocity formula v(m)=veln⁡(m0/m)=veln⁡(m0)−veln⁡(m)v(m) = v_e \ln(m_0 / m) = v_e \ln(m_0) - v_e \ln(m)v(m)=ve​ln(m0​/m)=ve​ln(m0​)−ve​ln(m) is differentiated to determine the vehicle's acceleration dvdt=−vemdmdt\frac{dv}{dt} = -\frac{v_e}{m}\frac{dm}{dt}dtdv​=−mve​​dtdm​ during fuel burn (see Rocket Velocity Optimization via Rocket Equation (Aerospace Engineering)).
  • Chemical Engineering (Acidity and pH Sensors): The pH of chemical solutions is tracked via pH=−log⁡10(H+)\text{pH} = -\log_{10}(H^+)pH=−log10​(H+). Calculating rate of pH change d(pH)dt=−1H+ln⁡(10)dH+dt\frac{d(\text{pH})}{dt} = -\frac{1}{H^+ \ln(10)}\frac{dH^+}{dt}dtd(pH)​=−H+ln(10)1​dtdH+​ provides feedback for automated buffering valves (see Reaction Rate Kinematics and pH Buffering (Chemical Engineering)).
  • Computer Science & Telecommunications (Entropy and Information): Shannon entropy and signal attenuation in communication channels depend on base-2 and natural logarithms.

Example Problems

Example 1: Differentiating with Pre-Processing

Differentiate h(x)=ln⁡(x2−xx2+2x)h(x) = \ln\left(\frac{x^2 - x}{x^2 + 2x}\right)h(x)=ln(x2+2xx2−x​) for x>1x > 1x>1.

Solution

Step 1: Simplify the argument algebraically Factor xxx from numerator and denominator:

x2−xx2+2x=x(x−1)x(x+2)=x−1x+2\frac{x^2 - x}{x^2 + 2x} = \frac{x(x - 1)}{x(x + 2)} = \frac{x - 1}{x + 2}x2+2xx2−x​=x(x+2)x(x−1)​=x+2x−1​

Step 2: Expand using the Quotient Property of logarithms

h(x)=ln⁡(x−1x+2)=ln⁡(x−1)−ln⁡(x+2)h(x) = \ln\left(\frac{x - 1}{x + 2}\right) = \ln(x - 1) - \ln(x + 2)h(x)=ln(x+2x−1​)=ln(x−1)−ln(x+2)

Step 3: Differentiate term-by-term using ddx[ln⁡(u)]=u′u\frac{d}{dx}[\ln(u)] = \frac{u'}{u}dxd​[ln(u)]=uu′​

h′(x)=1x−1−1x+2h'(x) = \frac{1}{x - 1} - \frac{1}{x + 2}h′(x)=x−11​−x+21​

Step 4: Combine over a common denominator

h′(x)=(x+2)−(x−1)(x−1)(x+2)=x+2−x+1(x−1)(x+2)=3(x−1)(x+2)h'(x) = \frac{(x + 2) - (x - 1)}{(x - 1)(x + 2)} = \frac{x + 2 - x + 1}{(x - 1)(x + 2)} = \frac{3}{(x - 1)(x + 2)}h′(x)=(x−1)(x+2)(x+2)−(x−1)​=(x−1)(x+2)x+2−x+1​=(x−1)(x+2)3​

Example 2: Product Rule with General Base Logarithm

Find the derivative of g(x)=3x2log⁡3(x)g(x) = 3x^2 \log_3(x)g(x)=3x2log3​(x).

Solution

Step 1: Identify the Product Rule structure Let u=3x2u = 3x^2u=3x2 and v=log⁡3(x)v = \log_3(x)v=log3​(x).

u′=6x,v′=1xln⁡(3)u' = 6x, \qquad v' = \frac{1}{x \ln(3)}u′=6x,v′=xln(3)1​

Step 2: Assemble the Product Rule

g′(x)=u′v+uv′=(6x)log⁡3(x)+(3x2)(1xln⁡(3))g'(x) = u'v + uv' = (6x)\log_3(x) + (3x^2)\left(\frac{1}{x \ln(3)}\right)g′(x)=u′v+uv′=(6x)log3​(x)+(3x2)(xln(3)1​)

Step 3: Simplify algebraic factors

g′(x)=6xlog⁡3(x)+3xln⁡(3)g'(x) = 6x \log_3(x) + \frac{3x}{\ln(3)}g′(x)=6xlog3​(x)+ln(3)3x​


Connections & References

  • Parent Concepts: Derivatives, Logarithmic and Exponential Properties
  • Prerequisites: Negative and Rational Exponents, Chain Rule
  • Inverse Companion: Derivatives of Exponential Functions
  • Advanced Continuation: Logarithmic Differentiation
  • Linked Engineering Applications: Rocket Velocity Optimization via Rocket Equation (Aerospace Engineering), Reaction Rate Kinematics and pH Buffering (Chemical Engineering)
  • Course Sessions: Calc-1 Session 08, Calc-1 Session 09