Powerful Trick for Taking Derivatives

Quickly Derive Anything

A remarkably simple and general rule for taking the derivative of a function, no matter how long and complicated, reads

$\begin{equation} \boxed{ \frac{d}{d x } \left[ f(x)^ag(x)^b\ldots \right] = \left( f(x)^ag(x)^b\ldots \right) \times \left( \frac{a}{f(x)}f'(x) + \frac{b}{g(x)}g'(x) + \ldots \right) } \end{equation}$

In words, simply write down the function you want to derive verbatim,

$\begin{equation} \frac{d}{d x } \left[ x^3\cos(7x^2)e^{2x} \right] = \left( x^3\cos(7x^2)e^{2x} \right) \times ( \end{equation}$

and then for each factor, add a term consisting of its exponent divided by the factor, times its derivative. The example has three factor functions, the first, $x^3$

$\begin{equation} \frac{d}{d x } \left[ x^3\cos(7x^2)e^{2x} \right] = \left( x^3\cos(7x^2)e^{2x} \right) \times ( \frac{3}{x} + \end{equation}$

the second, $\cos(7x^2)$

$\begin{equation} \frac{d}{d x } \left[ x^3\cos(7x^2)e^{2x} \right] = \left( x^3\cos(7x^2)e^{2x} \right) \times ( \frac{3}{x} - \frac{1}{\cos(7x^2)}14x\sin(7x^2) + \end{equation}$

and the third, $e^{2x}=(e^x)^2$ ,

$\begin{equation} \frac{d}{d x } \left[ x^3\cos(7x^2)e^{2x} \right] = \left( x^3\cos(7x^2)e^{2x} \right) \times \left( \frac{3}{x} - \frac{1}{\cos(7x^2)}14x\sin(7x^2) + \frac{2}{e^x}e^x \right). \end{equation}$

This simplifies to

$\begin{align} \frac{d}{d x } \left[ x^3\cos(7x^2)e^{2x} \right] &= \left( x^3\cos(7x^2)e^{2x} \right) \times \left( \frac{3}{x} - 14x \frac{\sin(7x^2)}{\cos(7x^2)} + 2 \right) \\ &= e^{2x}x^2 \left((2x+3)\cos(7x^2) - 14x^2\sin(7x^2) \right). \end{align}$

You are done! With this you can take the derivative of any function you are likely to encounter in less than a minute. This works nicely for large and complicated functions, unlike the rules taught in middle school. I found this trick in Richard Feynman's lectures, in notes written by Margaret Keck.

Proof

Let $f, g, \ldots$ be functions of $x$ , and let $a, b,\ldots$ be constants. Using the properties of the logarithm,

$\begin{align} \frac{d}{d x } \log(f^ag^b\ldots) &= \frac{d}{d x } \left[ a\log(f) + b\log(g) + \ldots \right] \\ &= \frac{a}{f}f' + \frac{b}{g}g' + \ldots \end{align}$

and

$\begin{equation} \frac{d}{d x } \log(f^ag^b\ldots) = \frac{\frac{d}{d x } \left[ f^ag^b\ldots \right]}{f^ag^b\ldots} \end{equation}$

From which it follows that

$\begin{equation} \frac{d}{d x } \left[ f^ag^b\ldots \right] = \left( f^ag^b\ldots \right) \left( \frac{a}{f}f' + \frac{b}{g}g' + \ldots \right) \quad\blacksquare \end{equation}$