Tropic of Calculus » Lesson 2.7: Exponential and Logarithmic Derivatives

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Lesson 2.7: Exponential and Logarithmic Derivatives

Two more categories of functions for which we must find derivatives are logarithmic and exponential functions. These derivatives are particularly useful as we continue in our study of calculus.

One way to find the derivative of $f(x)=e^x$ would simply be to use the definition of $e$ : $e$ is the number such that $\frac{d}{dx}(e^x)=e^x$ . This is entirely correct, of course, but it’s a little lame as a “proof”. We can quickly give a better proof by considering . We know that $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ and we can plug in $f$ : $f'(x)=\lim_{h\to0}\frac{e^{x+h}-e^x}{h}$

One of the laws of exponents now allows us to expand the first term of the limit’s numerator: $f'(x)=\lim_{h\to0}\frac{e^xe^h-e^x}{h}$ Then we can factor the numerator: $f'(x)=\lim_{h\to0}\frac{e^x(e^h-1)}{h}$ Since $e^x$ doesn’t depend on $h$ , we can remove it from the limit: $f'(x)=e^x\lim_{h\to0}\frac{e^h-1}{h}$ Graph of y=(e^h-1)/h We could use a tedious delta-epsilon procedure to evalute the remaining limit, but let’s just look at the graph of $y=\frac{e^h-1}{h}$ . It’s obvious that the limit as $x\to0$ is $1$ . (If you don’t believe it, the graph of is provided to the right for your convenience.) Therefore $f'(x)=e^x(1)=e^x$ ; that is, $\frac{d}{dx}(e^x)=e^x$ , just as we said.

We can use this definition to find a more general derivative for $b^x$ for some constant $b$ . $g(x)=b^x$ can be rewritten as $g(x)=e^{x\ln b}$ . Then we apply the chain rule and find that $g'(x)=e^{x\ln b}\cdot\ln b$ . We can apply the reverse of that rewriting and get $g'(x)=b^x\ln b$ .

To differentiate logarithmic functions, we again start with base $e$ . This time, we use implicit differentiation. Considering $y=\ln x$ we can raise $e$ to the power of both sides, getting $e^y=e^{\ln x}$ —or, rewriting the right side, $e^y=x$ . Applying implicit differentiation, $e^y\frac{dy}{dx}=1$ . Since $e^y=x$ , that equation is the same as $x\frac{dy}{dx}=1$ , so we solve for $\frac{dy}{dx}$ : $\frac{dy}{dx}=\frac1x$

Let’s generalize this to logs base ; that is, we want to find d/dx (log_b x) . Recall our old friend from Algebra II, the change-of-base formula: log_b x = (ln x)/(ln b) So we want to find d/dx (ln x/ln b) , and since 1/ln b does not depend on we can put it outside the derivative: d/dx (log_b x)=(1/ln b) d/dx (ln x) Since we know d/dx (ln x) = 1/x , we have the general formula for the derivative of a logarithm: d/dx (log_b x)=1/(x ln b)