# How do you find the derivative of y=e^(e^(3x^2))?

Oct 24, 2016

$y ' = 6 x {e}^{{e}^{3 {x}^{2}} + 3 {x}^{2}}$

#### Explanation:

The rule for differentiating e functions is,

$y = {e}^{f} \left(x\right)$

$y ' = f ' \left(x\right) {e}^{f} \left(x\right)$

so in the case of your question we have

$y = {e}^{f} \left(x\right)$ where $f \left(x\right) = {e}^{g} \left(x\right)$

so we will differentiate to find each part we need,
$y = {e}^{f} \left(x\right)$

$f \left(x\right) = {e}^{g} \left(x\right)$

$f ' \left(x\right) = g ' \left(x\right) {e}^{g} \left(x\right)$

So now what is y, f(x) and g(x).

$y = {e}^{{e}^{3 {x}^{2}}}$

$f \left(x\right) = {e}^{3 {x}^{2}}$

$g \left(x\right) = {e}^{3 {x}^{2}}$

$f \left(x\right) = {e}^{3 {x}^{2}}$

$f ' \left(x\right) = 6 x \cdot {e}^{3 {x}^{2}}$

so subbing into the original formula,

$y ' = f ' \left(x\right) {e}^{f} \left(x\right)$

$y ' = \left(6 x \cdot {e}^{3 {x}^{2}}\right) \left({e}^{{e}^{3 {x}^{2}}}\right)$

$y ' = 6 x \cdot {e}^{{e}^{3 {x}^{2}}} \cdot {e}^{3 {x}^{2}}$

$y ' = 6 x {e}^{{e}^{3 {x}^{2}} + 3 {x}^{2}}$