# How do you differentiate f(x)=e^(5x^2+x+3)  using the chain rule?

##### 2 Answers
Feb 25, 2016

$f ' \left(x\right) = \left(10 x + 1\right) {e}^{5 {x}^{2} + x + 3}$

#### Explanation:

Using the $\textcolor{b l u e}{\text{ chain rule }}$

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right)$

and $\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$

f'(x)$= {e}^{5 {x}^{2} + x + 3} \frac{d}{\mathrm{dx}} \left(5 {x}^{2} + x + 3\right)$

$= {e}^{5 {x}^{2} + x + 3} \left(10 x + 1\right)$

Feb 25, 2016

$h ' \left(x\right) = {e}^{5 {x}^{2} + x + 3} \left(10 x + 1\right)$

#### Explanation:

The given equation is $h \left(x\right) = {e}^{5 {x}^{2} + x + 3}$. But it seems as if it is a function inside a function. So, let's write $g \left(x\right) = 5 {x}^{2} + x + 3$ and $f \left(x\right) = {e}^{x}$.
So that means f(g(x))=e^(g(x))=e^(5x^2+x+3

From chain rule, we have d/dx(f(g(x))=h'(x)=f'(g(x))* g'(x).
So, taking for $f \left(g \left(x\right)\right)$ and $g \left(x\right)$, we get
$h ' \left(x\right) = {e}^{g \left(x\right)} \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right) = = {e}^{g} \left(x\right) \left(10 x + 1\right)$

You can substitute it all to get back the proper answer.