Can someone please explain how the power rule, brings natural log into the equation? Why is the derivative not $\arctan(x)(3x^2+5)^{\arctan(x)-1}$?

  • $\begingroup$ Your final expression is a bit unclear. Do you mean $\arctan(x)(3x^2+5)^{\arctan(x)-1}$? $\endgroup$– Chee HanApr 29, 2017 at 2:32
  • $\begingroup$ yes, sorry, I still have a hard time writing exponents on this site. $\endgroup$– user440071Apr 29, 2017 at 2:34

3 Answers

You can write $a^b$ as $e^{b \ln a}$. So in this case we have $$(3x^2 + 5)^{\arctan x} = \exp(\arctan x \ln (3x^2+5))$$ and we know that the derivative of $\exp f(x)$ is $f'(x) \exp f(x)$ by the chain rule.

As for why the derivative of $f^g$ is not simply $gf^{g-1}$ for differentiable functions $f$ and $g$ in general, well – it’s simply not true. We can provide many counterexamples. The ‘power rule’ only works if want to differentiate something of the form $x^c$ where $c$ is emphatically a constant.

  • $\begingroup$ okay so this rule would be used for any function with another function as it’s exponent? and similarly if sinx or cosx would be the exponent? $\endgroup$– user440071Apr 29, 2017 at 2:42
  • $\begingroup$ @M.Bucciacchio Yes, this is the de facto way of differentiating functions of the form $f^g$ where both $f,g$ are functions. You’d repeat the exact same process if the exponent were $\cos$ or $\sin$. $\endgroup$ Apr 29, 2017 at 2:43
  • $\begingroup$ No problem! $\,\,$ $\endgroup$ Apr 29, 2017 at 2:45

Recall that $e^{\log(x)}=x$. Hence, $$f(x)^{g(x)}=e^{\log(f(x)^{g(x)})}=e^{g(x)\log(f(x))}\tag 1$$

Using the chain rule in $(1)$, we can assert that

$$\frac{df(x)^{g(x)}}{dx}=e^{g(x)\log(f(x)}\frac{d(g(x)\log(f(x))}{dx}\tag 2$$

Then, let $f(x)=3x^2+5$ and $g(x)=\arctan(x)$ in $(2)$ and proceed.

For this kind of monsters, logarithmic differentiation is your best friend.

Making the problem more general, consider $$y={f(x)}^{g(x)}\implies \log(y)=g(x) \log(f(x))$$ Differentiate both sides $$\frac {y’}y=g(x)\frac{f'(x)}{f(x)}+g'(x)\log\left({f(x)} \right)$$ Now, use $$y’=y\times \frac {y’}y$$

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