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

##### 1 Answer
Dec 1, 2017

$\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = - {e}^{\sqrt{x}} / \left(2 \sqrt{x}\right)$

#### Explanation:

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

So,

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

As $g \left(x\right) = \sqrt{x}$

Therefore,

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

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 $\to$

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

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

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

$\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = - {e}^{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}$

Therefore,

$\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = - {e}^{\sqrt{x}} / \left(2 \sqrt{x}\right)$