# How do you differentiate y= e^(2x) (x^2 - 4) ln x?

Jul 30, 2015

#### Answer:

$y = {e}^{2 x} / x \cdot \left[2 \ln x \cdot \left({x}^{3} + {x}^{2} - 4 x\right) + {x}^{2} - 4\right]$

#### Explanation:

Start by rewriting your function as

$y = {e}^{2 x} \cdot {x}^{2} \cdot \ln x - 4 \cdot {e}^{2 x} \cdot \ln x$

You can use the sum rule to write

$\frac{d}{\mathrm{dx}} \left(y\right) = {\underbrace{\frac{d}{\mathrm{dx}} \left({e}^{2 x} \cdot {x}^{2} \cdot \ln x\right)}}_{\textcolor{red}{\frac{d}{\mathrm{dx}} \left(f\right)}} - 4 \cdot {\underbrace{\frac{d}{\mathrm{dx}} \left({e}^{2 x} \cdot \ln x\right)}}_{\textcolor{red}{\frac{d}{\mathrm{dx}} \left(g\right)}}$

Let's start with the second one because it is easier to calculate. For functions that can be written as the product of two other functions,

$y = f \left(x\right) \cdot g \left(x\right)$,

the product rule will allow you to calculate their derivative by

color(blue)(d/dx(y) = f^'(x) * g(x) + f(x) * g^'(x)

In your case, you can calculate $\frac{d}{\mathrm{dx}} \left({e}^{2 x} \cdot \ln x\right) = \frac{d}{\mathrm{dx}} \left(g\right)$ by

${g}^{'} = \frac{d}{\mathrm{dx}} \left({e}^{2 x}\right) \cdot \ln x + {e}^{2 x} \cdot \frac{d}{\mathrm{dx}} \left(\ln x\right)$

You can calculate $\frac{d}{\mathrm{dx}} \left({e}^{2 x}\right)$ by using the chain rule, for

$z \left(u\right) = {e}^{u}$ and $u = 2 x$, so that

$\frac{d}{\mathrm{dx}} \left(z\right) = \frac{d}{\mathrm{du}} \left(z\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left(z\right) = \frac{d}{\mathrm{du}} {e}^{u} \cdot \frac{d}{\mathrm{dx}} \left(2 x\right)$

$\frac{d}{\mathrm{dx}} \left(z\right) = {e}^{u} \cdot 2 = 2 \cdot {e}^{2 x}$

This means that you have

${g}^{'} = 2 \cdot {e}^{2 x} \cdot \ln x + {e}^{2 x} \cdot \frac{1}{x}$

${g}^{'} = {3}^{2 x} \left(2 \ln x + \frac{1}{x}\right)$

Now move to $\frac{d}{\mathrm{dx}} \left(f\right)$, which can be written as the product of 3 functions.

The general form of the product rule for such functions is

$\frac{d}{\mathrm{dx}} \left(a \cdot b \cdot c\right) = \frac{d}{\mathrm{dx}} \left(a \cdot b\right) \cdot c + \left(a \cdot b\right) \cdot \frac{d}{\mathrm{dx}} \left(c\right)$

$\frac{d}{\mathrm{dx}} \left(a \cdot b \cdot c\right) = \left[\frac{d}{\mathrm{dx}} \left(a\right) \cdot b + a \cdot \frac{d}{\mathrm{dx}} \left(b\right)\right] \cdot c + \left(a \cdot b\right) \cdot \frac{d}{\mathrm{dx}} \left(c\right)$

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(a \cdot b \cdot c\right) = \left[\frac{d}{\mathrm{dx}} \left(a\right)\right] \cdot b c + a \left[\frac{d}{\mathrm{dx}} \left(b\right)\right] c + a b \left[\frac{d}{\mathrm{dx}} \left(c\right)\right]}$

So, in your case, you have

${f}^{'} = \left[\frac{d}{\mathrm{dx}} \left({e}^{2 x}\right)\right] \cdot {x}^{2} \cdot \ln x + {e}^{2 x} \cdot \left[\frac{d}{\mathrm{dx}} \left({x}^{2}\right)\right] \cdot \ln x + {e}^{2 x} \cdot {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left(\ln x\right)$

${f}^{'} = 2 \cdot {e}^{2 x} \cdot {x}^{2} \cdot \ln x + 2 x \cdot {e}^{2 x} \cdot \ln x + {e}^{2 x} \cdot {x}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}$

${f}^{'} = x \cdot {e}^{2 x} \left(2 x \ln x + 2 \ln x + 1\right)$

Your target derivative will thus be

${y}^{'} = {f}^{'} - 4 {g}^{'}$

${y}^{'} = 2 \cdot {e}^{2 x} \cdot {x}^{2} \cdot \ln x + 2 x \cdot {e}^{2 x} \cdot \ln x + {e}^{2 x} \cdot x - 4 \cdot \left(2 \cdot {e}^{2 x} \cdot \ln x + {e}^{2 x} / x\right)$

${y}^{'} = 2 \cdot {e}^{2 x} \cdot {x}^{2} \cdot \ln x + 2 x \cdot {e}^{2 x} \cdot \ln x + {e}^{2 x} \cdot x - 8 {e}^{2 x} \cdot \ln x - \frac{4}{x} {e}^{2 x}$

${y}^{'} = {e}^{2 x} / x \cdot \left(2 {x}^{3} \cdot \ln x + 2 {x}^{2} \cdot \ln x + {x}^{2} - 8 x \cdot \ln x - 4\right)$

Finally, you get

${y}^{'} = \textcolor{g r e e n}{{e}^{2 x} / x \cdot \left[2 \ln x \cdot \left({x}^{3} + {x}^{2} - 4 x\right) + {x}^{2} - 4\right]}$

SIDE NOTE You could just as easily have used the product rule for three functions for your starting function, which would have resulted in

${y}^{'} = \left[\frac{d}{\mathrm{dx}} \left({e}^{2 x}\right)\right] \cdot \left({x}^{2} - 4\right) \cdot \ln x + {e}^{2 x} \cdot \left[\frac{d}{\mathrm{dx}} \left({x}^{2} - 4\right)\right] \cdot \ln x + {e}^{2 x} \cdot \left({x}^{2} - 4\right) \cdot \frac{d}{\mathrm{dx}} \left(\ln x\right)$