Question

How do you find $f'\left( x \right)$ and $f''\left( x \right)$ given $f\left( x \right)={{x}^{4}}{{e}^{x}}$

Answer 1

First of all we have to calculate $f'\left( x \right)$, which is a first order derivative. To find the value of $f'\left( x \right)$ we will use the product rule of differentiation which is given by
$\dfrac{d}{dx}\left[ f(x)g(x) \right]=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)$
Then to calculate $f''\left( x \right)$ we will differentiate the obtained value of $f'\left( x \right)$ by using the same product rule.

Complete step by step answer:
We have been given that $f\left( x \right)={{x}^{4}}{{e}^{x}}$
We have to find the values of $f'\left( x \right)$ and $f''\left( x \right)$.
Now, the given function $f\left( x \right)={{x}^{4}}{{e}^{x}}$ is the product of two functions and we have to find the derivative.
So we know that product rule to find the derivative of two functions is given by $\dfrac{d}{dx}\left[ f(x)g(x) \right]=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)$
Now, we have $f(x)={{x}^{4}}$ and $g(x)={{e}^{x}}$
Substituting the values in the above formula we get
$\Rightarrow \dfrac{d}{dx}\left[ {{x}^{4}}{{e}^{x}} \right]={{x}^{4}}\dfrac{d}{dx}{{e}^{x}}+{{e}^{x}}\dfrac{d}{dx}{{x}^{4}}$
Now, we know that by chain rule $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}\dfrac{d}{dx}x$
Substituting the values in the above equation we get
$\Rightarrow \dfrac{d}{dx}\left[ {{x}^{4}}{{e}^{x}} \right]={{x}^{4}}{{e}^{x}}\dfrac{d}{dx}x+{{e}^{x}}\dfrac{d}{dx}{{x}^{4}}$
Now, we know that $\dfrac{d}{dx}{{x}^{n}}=n\times {{x}^{n-1}}$
Substituting the values we get
$\Rightarrow \dfrac{d}{dx}\left[ {{x}^{4}}{{e}^{x}} \right]={{x}^{4}}{{e}^{x}}+{{e}^{x}}4{{x}^{4-1}}$
Now, solving further we get
$\Rightarrow \dfrac{d}{dx}\left[ {{x}^{4}}{{e}^{x}} \right]={{x}^{4}}{{e}^{x}}+4{{e}^{x}}{{x}^{3}}$
Hence we get the value of $f'\left( x \right)={{x}^{4}}{{e}^{x}}+4{{e}^{x}}{{x}^{3}}$
Now, to find the $f''\left( x \right)$ we need to differentiate the obtained function again.
We have $f'\left( x \right)={{x}^{4}}{{e}^{x}}+4{{e}^{x}}{{x}^{3}}$
Now, differentiating the above equation with respect to x we get
$\Rightarrow f''\left( x \right)=\dfrac{d}{dx}{{x}^{4}}{{e}^{x}}+\dfrac{d}{dx}4{{e}^{x}}{{x}^{3}}$
Now, we know that $\dfrac{d}{dx}\left[ f(x)g(x) \right]=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)$
Substituting the values we get
$\Rightarrow f''\left( x \right)={{e}^{x}}\dfrac{d}{dx}{{x}^{4}}+{{x}^{4}}\dfrac{d}{dx}{{e}^{x}}+4\left[ {{x}^{3}}\dfrac{d}{dx}{{e}^{x}}+{{e}^{x}}\dfrac{d}{dx}{{x}^{3}} \right]$
Now, we know that by chain rule $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}\dfrac{d}{dx}x$
Substituting the values we get
$\Rightarrow f''\left( x \right)={{e}^{x}}\dfrac{d}{dx}{{x}^{4}}+{{x}^{4}}{{e}^{x}}+4\left[ {{x}^{3}}{{e}^{x}}+{{e}^{x}}\dfrac{d}{dx}{{x}^{3}} \right]$
Now, we know that $\dfrac{d}{dx}{{x}^{n}}=n\times {{x}^{n-1}}$
Substituting the values we get
$\Rightarrow f''\left( x \right)=4{{e}^{x}}{{x}^{4-1}}+{{x}^{4}}{{e}^{x}}+4\left[ {{x}^{3}}{{e}^{x}}+3{{e}^{x}}{{x}^{3-1}} \right]$
Now, simplifying further we get

& \Rightarrow f''\left( x \right)=4{{e}^{x}}{{x}^{3}}+{{x}^{4}}{{e}^{x}}+4\left[ {{x}^{3}}{{e}^{x}}+3{{e}^{x}}{{x}^{2}} \right] \\\ & \Rightarrow f''\left( x \right)=4{{e}^{x}}{{x}^{3}}+{{x}^{4}}{{e}^{x}}+4{{x}^{3}}{{e}^{x}}+12{{e}^{x}}{{x}^{2}} \\\ & \Rightarrow f''\left( x \right)={{x}^{4}}{{e}^{x}}+8{{e}^{x}}{{x}^{3}}+12{{e}^{x}}{{x}^{2}} \\\ \end{aligned}$$ Hence we get the values of $f'\left( x \right)$ and $f''\left( x \right)$ as $f'\left( x \right)={{x}^{4}}{{e}^{x}}+4{{e}^{x}}{{x}^{3}}$ and $$f''\left( x \right)={{x}^{4}}{{e}^{x}}+8{{e}^{x}}{{x}^{3}}+12{{e}^{x}}{{x}^{2}}$$ **Note:** Avoid basic calculation mistakes. The point to remember about this question is that product rule and chain rule are different. Chain rule is applied when we have a composite function $f\left[ g(x) \right]$. Product rule is used when we have a product of two functions.