Question

How do you find the first and second derivative of ${{e}^{{{x}^{2}}}}$ ?

Answer 1

The given function is of the form ${{e}^{f\left( x \right)}}$ . So to find the derivative of the given form we have a standard form which we can follow. Now, we can differentiate it again to get the second derivative of the same function using the product rule for differentiation given by $\dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=f\left( x \right).g'\left( x \right)+g\left( x \right).f'\left( x \right)$ .

Complete step by step solution:
The function to find the derivative is ${{e}^{{{x}^{2}}}}$ .
Firstly we will find the first derivative of the given function as this function is of form ${{e}^{f\left( x \right)}}$ .
We know the standard form of the derivatives as follows, $\dfrac{d}{dx}\left( {{e}^{f\left( x \right)}} \right)={{e}^{f\left( x \right)}}.f'\left( x \right)$ .
From this we can apply to given function,
$\Rightarrow \dfrac{d}{dx}\left( {{e}^{{{x}^{2}}}} \right)={{e}^{{{x}^{2}}}}.\dfrac{d}{dx}\left( {{x}^{2}} \right)$
$\Rightarrow \dfrac{d}{dx}\left( {{e}^{{{x}^{2}}}} \right)={{e}^{{{x}^{2}}}}.2x=2x{{e}^{{{x}^{2}}}}$
So the first derivative of the given function is obtained from the above solving as $2x{{e}^{{{x}^{2}}}}$
Now let us differentiate this function obtained from first derivative again which is in the form $\dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)$ to get the second derivative. We can make use of the product rule of differentiation. So the formula is as follows,
$\dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=f\left( x \right).g'\left( x \right)+g\left( x \right).f'\left( x \right)$
So from this second derivative of the given function can be obtained as follows,
$\Rightarrow \dfrac{d}{dx}\left( 2x{{e}^{{{x}^{2}}}} \right)=2\left( x.\dfrac{d}{dx}\left( {{e}^{{{x}^{2}}}} \right)+{{e}^{{{x}^{2}}}}.\dfrac{d}{dx}\left( x \right) \right)$
We can be simplify it to get the form as,
$\Rightarrow \dfrac{d}{dx}\left( 2x{{e}^{{{x}^{2}}}} \right)=2\left( x.\left( 2x{{e}^{{{x}^{2}}}} \right)+{{e}^{{{x}^{2}}}}.1 \right)$
$\Rightarrow \dfrac{d}{dx}\left( 2x{{e}^{{{x}^{2}}}} \right)=2\left( 2{{x}^{2}}{{e}^{{{x}^{2}}}}+{{e}^{{{x}^{2}}}} \right)=\left( 4{{x}^{2}}{{e}^{{{x}^{2}}}}+2{{e}^{{{x}^{2}}}} \right)$
So we have got the second derivative of the function as $4{{x}^{2}}{{e}^{{{x}^{2}}}}+2{{e}^{{{x}^{2}}}}$ for the given function.
From this we can write that the first and second derivatives are $2x{{e}^{{{x}^{2}}}}$ and $4{{x}^{2}}{{e}^{{{x}^{2}}}}+2{{e}^{{{x}^{2}}}}$ respectively.

Note: To solve this problem one should be aware of the basic concepts of finding the derivatives of the functions in general or some standard forms. Any mistake in simplification can lead to the error in the answer. So one should be careful in answering these questions. In case of such functions, the derivative will always have the function itself in its terms. Here, we had ${{e}^{{{x}^{2}}}}$ , we can see that final derivatives also have the same function in them.