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Derivatives of Exponentials and Logarithms

April 9th, 2009 |

Author: Merrin

In this section, we will learn the derivatives of exponentials and logarithms.

$\cfrac{d}{dx} e^{x} =e^{x} \quad \quad \cfrac{d}{dx} \ln x=\cfrac{1}{x}$

$\cfrac{d}{dx}a^x=a^x\ln a \qquad \cfrac{d}{dx}\log_b{x}=\cfrac{1}{x\ln b}$

Be aware of the fact that e is the natural base where the top two formulas are “pretty” and do not contain other factors. We write ln x for the natural logarithm instead of

$\log _{b} x$

for an arbitrary base. The last two rules for the derivatives of an exponential with an arbitrary base and the derivatives of logarithms with an arbitrary base can be derived from the first two rules and the properties of exponentials and logarithms.

Example 1. Prove the arbitrary rules using the natural rules.

$\cfrac{d}{dx} a^{x} =a^{x} \ln a \quad \quad \cfrac{d}{dx} \log _{a} x=\cfrac{1}{x\ln a}$

Solution 1. To take the derivative of an arbitrary exponential, one technique is to write the expression in terms of the natural exponential and then use the chain rule.

$\begin{array}{l} {f(x)=a^{x} =e^{\ln a^{x} } =e^{x\ln a} } \\ {\cfrac{d}{dx} f(x)=\cfrac{d}{dx} e^{x\ln a} =e^{x\ln a} \ln a=a^{x} \ln a} \end{array}$

To find the derivative of an arbitrary logarithm, change the base to find

an equation in terms of the natural logarithm. There is a “chain” rule for logarithms or change of base formula which is

$\log _{a} b\log _{b} c=\log _{a} c$

Applying this formula we can find the derivative of an arbitrary logarithm.

${\log _{e} a\log _{a} x=\log _{e} x=\ln x} \\ {\log _{a} x=\cfrac{\ln x}{\ln a} } \\ {\cfrac{d}{dx} \log _{a} x=\cfrac{1}{x\ln a} }$

Example 2. Prove the first two new rules for the derivatives of exp(x) and ln x.

Solution 2. We can try to find the derivative of an exponential by directly applying the definition of the derivative.

$\cfrac{d}{dx}e^x= \mathop{\lim }\limits_{\Delta x\to 0}\cfrac{e^{x+\Delta x}-e^x}{\Delta x}= e^x \mathop{\lim }\limits_{\Delta x\to 0}\cfrac{ e^{\Delta x}-1}{\Delta x}=e^x \left( \mathop{\lim }\limits_{h\to 0}\cfrac{e^{h}-1}{h}\right)$

It turns out that limit in the parenthesis in equal to one which we will

prove shortly. Alternatively if you know the derivative of the natural logarithm

and the chain rule then you can find the derivative of the exponential function.

${f(x)=e^{x} } \\ {\ln f(x)=x} \\ {\cfrac{f'}{f} =1} \\f'=f\\ {\cfrac{d}{dx} e^{x} =e^{x} }$

Finding the derivative of the natural logarithm is a little tricky because the result is also expressed in terms of a limit. Starting with an arbitrary base we write.

${\cfrac{d}{dx} \log _{b} x} ={\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\log _{b} (x+\Delta x)-\log _{b} (x)}{\Delta x} =\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\log _{b} (1+\Delta x/x)}{\Delta x} } \\ ={\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{1}{x} \cfrac{\log _{b} (1+\Delta x/x)}{\Delta x/x} =\cfrac{1}{x} \mathop{\lim }\limits_{u\to 0} \cfrac{\log _{b} (1+u)}{u} } \\ = {\cfrac{1}{x} \mathop{\lim }\limits_{u\to 0} \log _{b} (1+u)^{1/u} =\cfrac{1}{x} \log _{b}\left( \mathop{\lim }\limits_{n\to \infty } \left(1+\cfrac{1}{n} \right)^{n}\right)}$

These two above limits are found to equal the number e by direct

computation.

$e = \mathop{\lim }\limits_{u\to 0} (1+u)^{1/u}=\mathop{\lim }\limits_{n\to \infty } \left(1+\cfrac{1}{n} \right)^{n}$

We can expand this quantity with the binomial theorem and then take n to

be infinite. This will give us an expression we use to find the actual value

of e.

$\left(1+n^{-1} \right)^{n}=1+n(n^{-1})+\cfrac{n(n-1)}{2!}(n^{-2})++\cfrac{n(n-1)(n-2)}{3!}(n^{-3})+\ldots$

We know how to deal with the limits of rational functions like these. Just

compare the leading coefficients.

$\mathop{\lim }\limits_{n\to \infty } \left(1+\cfrac{1}{n} \right)^{n}=\sum_{k=0}^{\infty}\cfrac{1}{k!}=2.7182818284590452354...=e$

This sum converges rapidly towards the value of e. Some marvel at the numerology of the repeating numbers 18281828 others just get on with their lives. We can go back and calculate the limit involved in the derivative of the

exponential function and show that it is one. First we need an expansion

to prove it.

$\mathop{\lim }\limits_{n\to \infty } \left(1+\cfrac{h}{n} \right)^n = \mathop{\lim }\limits_{n/h\to \infty } \left(1+\cfrac{h}{n} \right)^{\cfrac{n}{h}\displaystyle h} =e^h$

Again multiplying this out by the binomial theorem like before we find that

$e^h=1+h+\cfrac{1}{2!}h^2+\ldots$

Thus when the value of h tends to zero in the limit we have

$\mathop{\lim }\limits_{h\to 0} \cfrac{ e^{h}-1}{h}=1$

These limit expressions are numbers which can just be calculated. Don’t be

intimidated. Now, back to the derivative of the logarithm. Take the base also as e to get the natural logarithm then ln e=1. Notice then the coefficient goes away in the above expression for the derivative of the natural logarithm.

$\cfrac{d}{dx} \ln x=\cfrac{1}{x}\left(\mathop{\lim }\limits_{n\to \infty } \ln \left(1+\cfrac{1}{n} \right)^{n}\right) = \cfrac{1}{x}\ln e = \cfrac{1}{x}$

Example 3. Find the derivative of

$f(x)=x^{x}$

Solution 3. We don’t yet have a rule for the derivative of functions where the base and the exponent are both functions of x. By a slick trick, we can make x to the x look like a function we know how to differentiate with the chain rule.

$f(x)=x^{x} =c^{\ln (x^{x} )} =e^{x\ln x}$