Calculus I · Unit 2A · lesson

Logarithmic Differentiation

Concept

Learning objectives

Use logarithms to differentiate complicated products, quotients, and variable exponents.

Turn Products and Powers into Sums

Explanation

Before the formulas

Logarithmic differentiation is a structural rewrite. It is most useful when multiplication, division, roots, or variable exponents make direct differentiation unwieldy. Taking logarithms converts products to sums, quotients to differences, and exponents to coefficients, after which ordinary rules can see the structure clearly.

The logarithm introduces domain assumptions. Work on an interval where the original expression is nonzero, use lny\ln|y| when appropriate, and remember that differentiating the left side gives y/yy'/y. The final answer must be multiplied by the original function.

Logarithmic differentiation is useful because logs transform products into sums, quotients into differences, and exponents into coefficients before differentiation.
Read this graph as text

Logarithms turn difficult multiplication into easier addition. Logarithmic differentiation is useful because logs transform products into sums, quotients into differences, and exponents into coefficients before differentiation. The original expression contains a quotient, powers, a radical, and an exponential. Taking logarithms separates those structures. After expansion, each term can be differentiated with familiar rules. The final step, not shown in the last box, is to multiply by y and replace y with the original expression.

The visual uses labeled positions, solid and dashed line styles, and written descriptions so logarithms turn difficult multiplication into easier addition does not depend on color.

Why it matters: The visual should explain why logarithmic differentiation is a strategic rewrite rather than a mysterious special rule. It shows the structural simplification before any derivative is taken.

Visual study

Logarithmic differentiation is useful because logs transform products into sums, quotients into differences, and exponents into coefficients before differentiation.

Explanation

Logarithms turn products into sums and powers into multipliers

Some expressions are difficult because many factors or powers change at once. Taking logarithms reorganizes the structure: products become sums, quotients become differences, and exponents move to the front. Differentiation then becomes much more linear.

After differentiating lny\ln y, remember that the chain rule gives y/yy'/y. The final step is to multiply by the original yy, so keeping the original function visible saves time and reduces transcription errors.

Products, quotients, and variable powers can become much easier after taking logarithms. Logarithms convert multiplication into addition and exponents into coefficients, allowing ordinary differentiation rules to replace a forest of product rules.

The method is not limited to positive-looking formulas. It can often be applied on intervals using lny\ln|y|, provided the function does not cross zero there. The interval viewpoint keeps the algebra honest.

Logarithmic differentiation is useful when a function contains many multiplicative factors or when the variable appears in both the base and exponent.

Method

Logarithmic differentiation procedure

• Write y=y= the given positive expression, or use absolute values locally when appropriate. • Take ln\ln of both sides. • Expand products, quotients, and powers using log laws. • Differentiate implicitly. • Solve for yy', then replace yy with the original function.

Guided walkthrough

A product with variable powers

Differentiate

y=x3x2+1(x2)5.y=\frac{x^3\sqrt{x^2+1}}{(x-2)^5}.
Answer reveal

Worked solution

Write a real attempt before opening the supplied answer.

Common mistake

After differentiating lny\ln|y|, the left side is y/yy'/y, not merely yy'. Forgetting to multiply by yy at the end discards the original function.

Application

Output sensitivity in a multiplicative model

Suppose production is modeled by

Q=12K0.4L0.6,Q=12K^{0.4}L^{0.6},

where capital KK is fixed and labor LL varies. Taking logarithms gives

lnQ=ln12+0.4lnK+0.6lnL.\ln Q=\ln12+0.4\ln K+0.6\ln L.

Differentiating with respect to LL,

QQ=0.6L.\frac{Q'}{Q}=\frac{0.6}{L}.

Thus a 1%1\% increase in labor produces approximately a 0.6%0.6\% increase in output within the model. Logarithmic differentiation reveals relative sensitivity directly.

After the explanation

Use the section idea

Reading lens

Track which variable depends on which and use reciprocal or logarithmic structure only where its conditions hold.

Mental model

Implicit equations constrain variables together; inverse functions exchange inputs and outputs; logarithms turn products and powers into sums.

Decision

Choose implicit, inverse, or logarithmic differentiation from the equation's representation, not from surface complexity.

Common trap

Dropping a y-prime factor, using a reciprocal slope at the wrong point, or ignoring domain restrictions.

Check yourself

Can you identify the correspondence point and all hidden dependencies before differentiating?

Interactive checkapp-elasticity-01

For Q=12K0.4L0.6Q=12K^{0.4}L^{0.6} with KK fixed, find labor elasticity.

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Show hint

Compute LQL/QLQ_L/Q.

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Vocab
Derivative
Math glossaryDerivative
ddx[f(x)]\frac d{dx}[f(x)]dydx\frac{dy}{dx}

The instantaneous rate of change of a function.

Learn more
Derivative notations
Math glossaryDerivative notations
ddx[f(x)]\frac d{dx}[f(x)]dydx=f(x)\frac{dy}{dx}=f'(x)

Different notations emphasize the operator, dependent variable, function, or time.

Learn more
Chain rule
Math glossaryChain rule
ddxf(g(x))=f(g(x))g(x)\frac d{dx}f(g(x))=f'(g(x))g'(x)

Differentiates a composition from outside to inside.

Learn more
Product rule
Math glossaryProduct rule
(fg)=fg+fg(fg)'=f'g+fg'

Differentiates a product as two cross contributions.

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Tangent line
Math glossaryTangent line
yf(a)=f(a)(xa)y-f(a)=f'(a)(x-a)

A line matching a curve's instantaneous direction at a point.

Learn more
Math glossary