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The logarithm

20 一月, 2016 - 10:05
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The natural logarithm is the function that undoes the exponential. In a situation like this, we have

\frac{dy}{dx}=\frac{1}{dx/dy}

where on the left we're thinking of y as a function of x, and on the right we consider x to be a function of y. Applying this to the natural logarithm,

\begin{align*} y &= \textrm{In} x \\ x&=e^y \\ \frac{dx}{dy} &=e^y \\ \frac{dy}{dx}&=\frac{1}{e^y} \\ &= \frac{1}{x}\\ \frac{\textrm{dIn } x}{dx} &=\frac{1}{x} \end{align*}

media/image11.png
Figure 2.12  Differentiation and integration of functions of the form
 
x^n. Constants out in front of the functions are not shown, so keep in mind that, for example, the derivative of x^2 isn’t x, it’s 2x.

This is noteworthy because it shows that there must be an exception to the rule that the derivative of x^n is nx^{n-1}, and the integral of x^{n-1} is x^n/n. (In The chain rule I remarked that this rule could be proved using the product rule for negative integer values of k, but that I would give a simpler, less tricky, and more general proof later. The proof is Example below.) The integral of x^{-1} is not x^0/0, which wouldn't make sense anyway because it involves division by zero. 1 Likewise the derivative of x^0=1 is 0x^{-1}, which is zero. Figure 2.12 shows the idea. The functions xn form a kind of ladder, with differentiation taking us down one rung, and integration taking us up. However, there are two special cases where differentiation takes us of the ladder entirely.

Example

Prove d(x^n)/dx=nx^{n-1} for any real value of n, not just an integer.\begin{align*} y &=x^n \\ &= e^{\textrm{nInx}} \end{align*}
By the chain rule,\begin{align*} \frac{dy}{dx} &=e^{\textrm{nInx}}\cdot \frac{n}{x} \\ &=x^n\cdot \frac{n}{x} \\ &= nx^{n-1} \end{align*}
(For n=0, the result is zero.)

When I started the discussion of the derivative of the logarithm, I wrote y=\textrm{In} x right of the bat. That meant I was implicitly assuming x was positive. More generally, the derivative of \textrm{In}\left | x \right | equals 1/x, regardless of the sign (see Problem 2.29 ).

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