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Derivative of ex

11 November, 2015 - 10:43

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}