Derivative of ex

Available under Creative Commons-ShareAlike 4.0 International License.

All of the reasoning on page 39 would have applied equally well to any other exponential function with a different base, such as $2^{x}$ or $10^{x}$ . Those functions would have different values of c, so if we want to deter- mine the value of cfor the base-ecase, we need to bring in the definition of e, or of the exponential function $e^{x}$ , somehow.

We can take the definition of $e^{x}$ to be

$e^{x}=_{n\rightarrow \infty}^{lim }(1+\frac{x}{n})^{n}$

The idea behind this relation is similar to the idea of compound interest. If the interest rate is 10%, compounded annually, then x= 0.1, and the balance grows by a factor (1 + x) = 1.1 in one year. If, instead, we want to compound the interest monthly, we can set the monthly interest rate to 0.1/12, and then the growth of the balance over a year is $(1+x/12)^{12}$ = 1.1047, which is slightly larger because the interest from the earlier months itself accrues interest in the later months. Continuing this limiting process, we find $e^{1.1}$ = 1.1052.

If n is large, then we have a good approximation to the base-e exponential, so let’s differentiate this finite-napproximation and try to find an approximation to the derivative of $e^{x}$ . The chain rule tells is that the derivative of $(1+x/n)^{n}$ is the derivative of the raising-to- the-nth-power function, multiplied by the derivative of the inside stuff,

d(1 + x/n)/dx= 1/n. We then have

$\frac{d(1+\frac{x}{n})^{n}}{dx}=[n(1+\frac{x}{n})^{n-1}]\cdot \frac{1}{n}$ $=(1+\frac{x}{n})^{n-1}$

But evaluating this at x= 0 simply gives 1, so at x= 0, the approximation to the derivative is exactly 1 for all values of n— it’s not even necessary to imagine going to larger and larger values of n. This establishes that c= 1, so we have for all values of x.

$\frac{de^{x}}{dx}=e^{x}$

1894 reads

You are here

Derivative of ex