If f(x)= - e^x  and g(x) = 5 x , how do you differentiate f(g(x))  using the chain rule?

Dec 4, 2017

$- 5 {e}^{5 x}$

Explanation:

$\text{to evaluate "f(g(x)" substitute "g(x)" into } f \left(x\right)$

rArrf(g(x)=f(5x)=-e^(5x)

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right]$

$= - {e}^{5 x} \times \frac{d}{\mathrm{dx}} \left(5 x\right) = - 5 {e}^{5 x}$

Dec 4, 2017

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

Explanation:

Given $f \left(x\right) = - {e}^{x}$ and $g \left(x\right) = 5 x$

So,

$f \left(g \left(x\right)\right) = - {e}^{g \left(x\right)}$

As $g \left(x\right) = 5 x$

Therefore,

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

Now differentiate both sides with respect to $x$ using the chain rule.

The chain rule says that $\to$

$\left(U \left(V \left(x\right)\right)\right) ' = U ' \left(V \left(x\right)\right) \times V ' \left(x\right)$

Now back to the question $\to$

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

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

$\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = - {e}^{5 x} \times \frac{d}{\mathrm{dx}} 5 x$

$\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = - {e}^{5 x} \times 5$

Therefore,

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