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

Feb 8, 2016

f'(x) =( 2e^(4x)sec^2e^(4x))/sqrt(tane^(4x)

#### Explanation:

making use of the chain rule

 d/dx(f(g(x)) = f'(g(x)).g'(x)

and knowing $\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x \textcolor{b l a c k}{\text{ and }} \frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$

first step is to rewrite $\sqrt{\tan {e}^{4 x}} = {\left(\tan {e}^{4 x}\right)}^{\frac{1}{2}}$

$f ' \left(x\right) = \frac{1}{2} {\left(\tan {e}^{4 x}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(\tan {e}^{4 x}\right)$

now $\frac{d}{\mathrm{dx}} \left(\tan {e}^{4 x}\right) = {\sec}^{2} {e}^{4 x} \frac{d}{\mathrm{dx}} \left({e}^{4 x}\right) \ldots \ldots \ldots \ldots . \left(1\right)$

and $\frac{d}{\mathrm{dx}} \left({e}^{4 x}\right) = {e}^{4 x} \frac{d}{\mathrm{dx}} \left(4 x\right) = 4 {e}^{4 x} \ldots \ldots \ldots \ldots \ldots . \left(2\right)$

replacing the results of (1) and (2) back into f'(x)

$f ' \left(x\right) = \frac{1}{2} {\left(\tan {e}^{4 x}\right)}^{- \frac{1}{2}} . {\sec}^{2} {e}^{4 x} . 4 {e}^{4 x}$

and rewriting negative exponent as radical

 rArr f'(x) =( 2e^(4x)sec^2e^(4x))/sqrt(tane^(4x)