Question

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

Answer 1

To Solve the given question, we need to know some of the properties of differentiation, and derivatives of some functions. We should know the product rule of derivatives which states that $\dfrac{d\left( f(x)g(x) \right)}{dx}=\dfrac{d\left( f(x) \right)}{dx}g(x)+f(x)\dfrac{d\left( g(x) \right)}{dx}$. We should also know the derivatives of ${{x}^{2}}\And {{e}^{x}}$. The derivatives of these functions with respect to x are $2x\And {{e}^{x}}$ respectively.

Complete step by step solution:
We are asked to differentiate the function $y={{x}^{2}}{{e}^{{{x}^{2}}}}$. We can evaluate the derivative as,
$\dfrac{dy}{dx}=\dfrac{d\left( {{x}^{2}}{{e}^{{{x}^{2}}}} \right)}{dx}$
We can evaluate the $\dfrac{d\left( {{x}^{2}}{{e}^{{{x}^{2}}}} \right)}{dx}$ by using the product rule. The product rule of differentiation states that, $\dfrac{d\left( f(x)g(x) \right)}{dx}=\dfrac{d\left( f(x) \right)}{dx}g(x)+f(x)\dfrac{d\left( g(x) \right)}{dx}$. For this question, we have $f(x)={{x}^{2}}\And g(x)={{e}^{{{x}^{2}}}}$. Thus, using the product rule, we get $\dfrac{d\left( {{x}^{2}}{{e}^{{{x}^{2}}}} \right)}{dx}=\dfrac{d\left( {{x}^{2}} \right)}{dx}{{e}^{{{x}^{2}}}}+{{x}^{2}}\dfrac{d\left( {{e}^{{{x}^{2}}}} \right)}{dx}$.
Thus, we can evaluate the derivate of given expression as follows,

& \Rightarrow \dfrac{d\left( {{x}^{2}}{{e}^{{{x}^{2}}}} \right)}{dx}=\dfrac{d\left( {{x}^{2}} \right)}{dx}{{e}^{{{x}^{2}}}}+{{x}^{2}}\dfrac{d\left( {{e}^{{{x}^{2}}}} \right)}{dx} \\\ & \Rightarrow \dfrac{d\left( {{x}^{2}}{{e}^{{{x}^{2}}}} \right)}{dx}=2x{{e}^{{{x}^{2}}}}+{{x}^{2}}{{e}^{{{x}^{2}}}}(2x) \\\ & \Rightarrow \dfrac{d\left( {{x}^{2}}{{e}^{{{x}^{2}}}} \right)}{dx}=2x{{e}^{{{x}^{2}}}}+2{{x}^{3}}{{e}^{{{x}^{2}}}} \\\ \end{aligned}$$ This is the first derivative of the expression, we can simplify the above expression to express it as, $$\Rightarrow \dfrac{dy}{dx}=2{{e}^{{{x}^{2}}}}\left( x+{{x}^{3}} \right)$$ differentiating the above expression once, again by using the product rule, we get $$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( 2{{e}^{{{x}^{2}}}} \right)}{dx}\left( x+{{x}^{3}} \right)+2{{e}^{{{x}^{2}}}}\dfrac{d\left( \left( x+{{x}^{3}} \right) \right)}{dx}$$ $$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( 2\times 2x{{e}^{{{x}^{2}}}} \right)\left( x+{{x}^{3}} \right)+2{{e}^{{{x}^{2}}}}\left( 1+3{{x}^{2}} \right)$$ We can simplify the above expression to write it as, $$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \left( 2x \right)\left( x+{{x}^{3}} \right)+\left( 1+3{{x}^{2}} \right) \right)2{{e}^{{{x}^{2}}}}$$ Expanding the brackets in the above expression, we get $$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( 1+2{{x}^{4}}+5{{x}^{2}} \right)2{{e}^{{{x}^{2}}}}$$ This is the second derivative of the given function. **Note:** To Solve these types of questions, we should know the different properties/ rules of differentiation like product rule, quotient rule, etc. Also, we should know the derivatives of different functions. For this question, we used the derivatives of function $$\sin x$$, $${{x}^{2}}\And {{\tan }^{-1}}x$$. We should also know the different methods like substitution to evaluate the derivative of complex functions. We can also use the following method to differentiate the functions of these types. The given function is of the form $$y=f(x){{e}^{f(x)}}$$. The first derivative of these functions can be evaluated as $$\dfrac{dy}{dx}=\left( 1+f(x) \right){{e}^{f(x)}}\dfrac{d\left( f(x) \right)}{dx}$$