# How do you differentiate f(x)=sqrt(1-e^(4x)) using the chain rule.?

May 9, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{- 2 {e}^{4 x}}{\sqrt{1 - {e}^{4 x}}}$

#### Explanation:

Here we use the concept of function of a function and use chain rule.

Here we have $f \left(x\right) = \sqrt{g \left(x\right)}$, where g(x)=1-e^(h(x) and $h \left(x\right) = 4 x$.

According to chain rule $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}} \times \frac{\mathrm{dh}}{\mathrm{dx}}$

Hence $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1}{2 \sqrt{1 - {e}^{4 x}}} \times \left(- {e}^{4 x}\right) \times 4$

or $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{- 2 {e}^{4 x}}{\sqrt{1 - {e}^{4 x}}}$