# How do you differentiate  y =e^(2x )(x^2+5^x) using the chain rule?

Dec 8, 2015

$y ' = {e}^{2 x} \left(2 {x}^{2} + 2 \left({5}^{x}\right) + 2 x + {5}^{x} \ln 5\right)$

#### Explanation:

Through the product rule:

$y ' = \left({x}^{2} + {5}^{x}\right) \textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[{e}^{2 x}\right]} + {e}^{2 x} \textcolor{m a \ge n t a}{\frac{d}{\mathrm{dx}} \left[{x}^{2} + {5}^{x}\right]}$

Find each derivative individually (both will require the chain rule):

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[{e}^{2 x}\right]} = {e}^{2 x} \frac{d}{\mathrm{dx}} \left[2 x\right] = \textcolor{b l u e}{2 {e}^{2 x}}$

color(magenta)(d/dx[x^2+5^x])=2x+color(green)(d/dx[5^x]

Finding $\frac{d}{\mathrm{dx}} \left[{5}^{x}\right]$ will be much simpler if we recognize that ${5}^{x} = {e}^{\ln {5}^{x}} = {e}^{x \ln 5}$.

color(green)(d/dx[e^(xln5)])=overbrace(e^(xln5))^("still"=5^x)d/dx[xln5]=color(green)(5^xln5

So, plug this back into the derivative we were finding earlier:

d/dx[x^2+5^x]=color(magenta)(2x+5^xln5

Plug everything we know back into our original equation for $y '$:

$y ' = \textcolor{b l u e}{2 {e}^{2 x}} \left({x}^{2} + {5}^{x}\right) + {e}^{2 x} \left(\textcolor{m a \ge n t a}{2 x + {5}^{x} \ln 5}\right)$

$y ' = {e}^{2 x} \left(2 {x}^{2} + 2 \left({5}^{x}\right) + 2 x + {5}^{x} \ln 5\right)$