# What are the first and second derivatives of f(x)=5^((x^5)-9x)?

Jan 1, 2016

Following the rule to differentiate exponential functions: $\frac{d \left({a}^{b}\right)}{\mathrm{dx}} = {a}^{b} \left(\ln \left(a\right)\right) b '$, we can proceed.

#### Explanation:

We'll rename $u = {x}^{5} - 9 x$ so that $f \left(x\right) = {5}^{u}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {5}^{u} \left(\ln \left(5\right)\right) u '$

Substituting $u$ and $u '$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {5}^{{x}^{5} - 9 x} \ln \left(5\right) \left(5 {x}^{4} - 9\right)$

As for the second derivative, we must see we have a three terms product. To differentiate it, we'll consider two of them as one, in a sort of chain rule, and then derivate these two as well, as shown in the formula below:

$\left(a b c\right) ' = \left(a b\right) ' c + \left(a b\right) c ' = a ' b c + a b ' c + a b c '$

Let's just identify things here. Following the general formula:
$\textcolor{red}{a = {5}^{{x}^{5} - 9 x}}$, $\textcolor{b l u e}{b = \ln \left(5\right)}$, $\textcolor{g r e e n}{c = 5 {x}^{4} - 9}$

$\frac{{\mathrm{dy}}^{2}}{{d}^{2} x} = \textcolor{red}{{5}^{{x}^{5} - 9 x} \ln \left(5\right) \left(5 {x}^{4} - 9\right)} \textcolor{b l u e}{\left(\ln \left(5\right)\right)} \textcolor{g r e e n}{\left(5 {x}^{4} - 9\right)} + \cancel{\textcolor{red}{{5}^{{x}^{5} - 9 x}} \textcolor{b l u e}{\left(0\right)} \textcolor{g r e e n}{\left(5 {x}^{4} - 9\right)}} + \textcolor{red}{{5}^{{x}^{5} - 9 x}} \textcolor{b l u e}{\left(\ln \left(5\right)\right)} \textcolor{g r e e n}{\left(20 {x}^{3}\right)}$

As zero cancels the product of the second term, we have left:

$\frac{{\mathrm{dy}}^{2}}{{d}^{2} x} = \left({5}^{{x}^{5} - 9 x} \ln \left(5\right)\right) \left(\left(5 {x}^{4} - 9\right) \ln \left(5\right) + 20 {x}^{3}\right)$