# How do you find the derivative of f(x)=(e^(2x)- 3lnx)^4?

Jun 28, 2016

4 (2e^(2x)-(3/x))× (e^(2x)-3lnx)^3

#### Explanation:

The derivative of $f \left(x\right)$ can be calculated using chain rule that says:

$f \left(x\right)$ can be written as composite functions where:

$v \left(x\right) = {e}^{2 x} - 3 \ln x$
$u \left(x\right) = {x}^{4}$
So,
$f \left(x\right) = u \left(v \left(x\right)\right)$
Applying chain rule on the composite function $f \left(x\right)$we have:
color (purple)(f'(x)=u (v (x))'
color (purple)(f'(x)=v'(x)×u'(v (x)))

Let's find color (purple)(v'(x)
Applying chain rule on the derivative of exponential:
color (red)((e^(g (x)))'=g'(x)×e^(g (x)))
Knowing the derivative of $\ln \left(x\right)$ that says:
$\textcolor{b r o w n}{\left(\ln \left(g \left(x\right)\right)\right) ' = \frac{g ' \left(x\right)}{g \left(x\right)}}$
$\textcolor{p u r p \le}{v ' \left(x\right)} = \textcolor{red}{\left(2 x\right) ' {e}^{2 x}} - 3 \textcolor{b r o w n}{\frac{x '}{x}}$

$\textcolor{p u r p \le}{\left(v ' \left(x\right)\right) = 2 {e}^{2 x} - \left(\frac{3}{x}\right)}$

Let's find $\textcolor{b l u e}{u ' \left(x\right)}$:
Applying the derivative of power stated as follows:
color (green)(x^n=nx^(n-1)
$\textcolor{b l u e}{u ' \left(x\right)} = \textcolor{g r e e n}{4 {x}^{3}}$

Based on chain rule above we need $u ' \left(v \left(x\right)\right)$ so let's substitute $x$ by $v \left(x\right)$:
$u ' \left(v \left(x\right)\right) = 4 {\left(v \left(x\right)\right)}^{3}$
$\textcolor{p u r p \le}{u ' \left(v \left(x\right)\right) = 4 {\left({e}^{2 x} - 3 \ln x\right)}^{3}}$

Let's substitute the values of $u ' \left(v \left(x\right)\right)$and $v ' \left(x\right)$ in the above chain rule above we have:
color (purple)(f'(x)=v'(x)×u'(v (x)))
color (purple)(f'(x)=(2e^(2x)-(3/x))× 4 (e^(2x)-3lnx)^3)
color (purple)(f'(x)=4 (2e^(2x)-(3/x))× (e^(2x)-3lnx)^3)