# How do you differentiate f(x)=8e^(x^2)/(e^x+1) using the chain rule?

Feb 13, 2016

The only trick here is that $\left({e}^{{x}^{2}}\right) ' = {e}^{{x}^{2}} \cdot \left({x}^{2}\right) ' = {e}^{{x}^{2}} \cdot 2 x$
Final derivative is:

$f ' \left(x\right) = 8 {e}^{{x}^{2}} \frac{2 x \cdot \left({e}^{x} + 1\right) - {e}^{x}}{{e}^{x} + 1} ^ 2$
or
$f ' \left(x\right) = 8 {e}^{{x}^{2}} \frac{{e}^{x} \cdot \left(2 x - 1\right) + 2 x + 1}{{e}^{x} + 1} ^ 2$

#### Explanation:

$f \left(x\right) = 8 \frac{{e}^{{x}^{2}}}{{e}^{x} + 1}$

$f ' \left(x\right) = 8 \frac{\left({e}^{{x}^{2}}\right) ' \left({e}^{x} + 1\right) - {e}^{{x}^{2}} \left({e}^{x} + 1\right) '}{{e}^{x} + 1} ^ 2$

$f ' \left(x\right) = 8 \frac{{e}^{{x}^{2}} \cdot \left({x}^{2}\right) ' \left({e}^{x} + 1\right) - {e}^{{x}^{2}} \cdot {e}^{x}}{{e}^{x} + 1} ^ 2$

$f ' \left(x\right) = 8 \frac{{e}^{{x}^{2}} 2 x \cdot \left({e}^{x} + 1\right) - {e}^{{x}^{2}} \cdot {e}^{x}}{{e}^{x} + 1} ^ 2$

$f ' \left(x\right) = 8 \frac{{e}^{{x}^{2}} \left(2 x \cdot \left({e}^{x} + 1\right) - {e}^{x}\right)}{{e}^{x} + 1} ^ 2$

$f ' \left(x\right) = 8 {e}^{{x}^{2}} \frac{2 x \cdot \left({e}^{x} + 1\right) - {e}^{x}}{{e}^{x} + 1} ^ 2$

or (if you want to factor ${e}^{x}$ in the nominator)

$f ' \left(x\right) = 8 {e}^{{x}^{2}} \frac{{e}^{x} \cdot \left(2 x - 1\right) + 2 x + 1}{{e}^{x} + 1} ^ 2$

Note: if you want to study the sign, you are gonna have a bad time. Just look at the graph:

graph{8(e^(x^2))/(e^x+1) [-50.25, 53.75, -2.3, 49.76]}