Derivatives of Logs

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Rates of Change

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Derivatives of Trig Functions

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Version 2
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Implicit Differentiation

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Derivatives of Logs

Formulas and Examples
Logarithmic Differentiation

Derivatives in Science

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In Economics
In Biology

Related Rates

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Example (shadow)

Linear Approximation and Differentials

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An example with negative

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Differentiation Review

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Logarithmic Differentiation
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$\frac{0}{0}$ equal?
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The Fundamental Theorem of Calculus

Three Different Concepts
The Fundamental Theorem of Calculus (Part 2)
The Fundamental Theorem of Calculus (Part 1)
More FTC 1

Logarithmic Differentiation

Now that we know the derivative of a log, we can combine it with the chain rule:

$\frac{d}{dx}\Big( \ln(y)\Big)= \frac{1}{y} \frac{dy}{dx}.$ Sometimes it is easier to take the derivative of

$\ln(y)$ than of

$y$ , and it is the only way to differentiate some functions. This is called logarithmic differentiation.

The process of differentiating $y=f(x)$ with logarithmic differentiation is simple. Take the natural log of both sides, then differentiate both sides with respect to $x$ . Solve for $\frac{dy}{dx}$ and write $y$ in terms of $x$ and you are finished.

Consider the function

$(f(x))^{(g(x))}$ , for any (non-constant) functions

$f$ and

$g$ . This is a function for which we do not have a differentiation rule; it is not a power function (because there is an

$x$ in the exponent), nor is it an exponential function (because there is an

$x$ in the base). Whenever you wish to differentiate

$(f(x))^{(g(x))}$ , logarithmic differentiation works beautifully. This is because once we take logs, we can pull the power down and use the product rule.

Example: Find the derivative of

$y=x^x$ .

Solution: (Notice that both

$f(x)=x$ and

$g(x)=x$ above.) Take the log of both sides to get

$\ln(y) = \ln(x^x)=x \ln(x)$ (see how we can pull that exponent

$x$ down?). By the product rule, the derivative of

$x \ln(x)$ is

$\ln(x) + 1$ (do this work and simplify to get that answer). So we have

$\frac{1}{y}\frac{dy}{dx}=\ln(x) + 1$ and when we replace

$y$ with

$x^x$ and solve for

$\frac{dy}{dx}$ we get

$\frac{dy}{dx} = y \frac{d}{dx}\Big(\ln(y)\Big) = x^x (1+\ln(x)).$