If f(x)= (5x -1)^3-2  and g(x) = e^x , what is f'(g(x)) ?

Apr 13, 2018

$15 {e}^{x} {\left(5 {e}^{x} - 1\right)}^{2}$

Explanation:

First, let's find $f \left(g \left(x\right)\right)$

${\left(5 \left({e}^{x}\right) - 1\right)}^{3} - 2$

Now we take the derivative of this:

Power rule

$f ' \left(g \left(x\right)\right) = 3 {\left(5 {e}^{x} - 1\right)}^{2}$

We need to apply the power rule and take the derivative of what's inside the parentheses

$\left(5 {e}^{x} - 1\right) ' = 5 {e}^{x}$

So our final answer is $f ' \left(g \left(x\right)\right) = 3 {\left(5 {e}^{x} - 1\right)}^{2} \times 5 {e}^{x}$ or $15 {e}^{x} {\left(5 {e}^{x} - 1\right)}^{2}$

Apr 13, 2018

$\implies f ' \left(g \left(x\right)\right) = 15 {\left(5 {e}^{x} - 1\right)}^{2}$

Explanation:

$f \left(x\right) = {\left(5 x - 1\right)}^{3} - 2$

$f \left(g \left(x\right)\right) = {\left(5 {e}^{x} - 1\right)}^{3} - 2$

$f ' \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left[{\left(5 {e}^{x} - 1\right)}^{3} - 2\right]$

$\implies f ' \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} {\left(5 {e}^{x} - 1\right)}^{3} - \frac{d}{\mathrm{dx}} \left(2\right)$

$\implies f ' \left(g \left(x\right)\right) = 15 {e}^{x} {\left(5 {e}^{x} - 1\right)}^{2}$

Hence, the solution is:

$\implies f ' \left(g \left(x\right)\right) = 15 {e}^{x} {\left(5 {e}^{x} - 1\right)}^{2}$