Question

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

Answer 1

Hint: We can find the first and second derivative of ${{x}^{2}}{{e}^{x}}$ by using basic formulae of differentiation. We can use the product rule formula to find the first and second derivative of ${{x}^{2}}{{e}^{x}}$ Formula for the product rule is given as $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$

 Complete step by step answer:
From the given question it had been asked to find the first and second derivative of ${{x}^{2}}{{e}^{x}}$
By using the product rule formula shown above, we can get the derivatives,
By applying the product rule formula we get, $\dfrac{dy}{dx}\text{ = 2x }\times \text{ }{{\text{e}}^{x}}\text{ + }{{\text{e}}^{x}}\text{ }\times \text{ }{{\text{x}}^{2}}\text{ }$
We know that $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$ and $\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$
On furthermore simplifying we get, $\dfrac{dy}{dx}\text{ = 2x}{{\text{e}}^{x}}\text{ + }{{\text{e}}^{x}}{{\text{x}}^{2}}\text{ }$
Now, we have to take $x{{e}^{x}}$ common on both sides of the equation to make it simplified.
On taking $x{{e}^{x}}$ common on both sides of the equation, we get $\dfrac{dy}{dx}\text{ = }x\left( x+2 \right){{e}^{x}}\text{ }$
Therefore, we got the first derivative of the given question.
Now differentiate the first derivative to get the second derivative.
We can differentiate the first derivative by applying the product rule formula.
By applying the product rule formula for first derivative we get,
$\dfrac{dy}{dx}\text{ = }x\left( x+2 \right){{e}^{x}}\text{ }$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\text{ }\times \text{ }{{\text{e}}^{x}}\text{ + 2x }\times \text{ }{{\text{e}}^{x}}\text{ + 2x }\times \text{ }{{\text{e}}^{x}}\text{ + }{{\text{x}}^{2}}\text{ }\times \text{ }{{\text{e}}^{x}}$
On furthermore simplifying we get,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2{{\text{e}}^{x}}\text{ + 2x}{{\text{e}}^{x}}\text{ + 2x}{{\text{e}}^{x}}\text{ + }{{\text{x}}^{2}}{{\text{e}}^{x}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2{{\text{e}}^{x}}\text{ + 4x}{{\text{e}}^{x}}\text{ + }{{\text{x}}^{2}}{{\text{e}}^{x}}$
Now, take ${{e}^{x}}$ common on the right hand side of the equation.
By taking ${{e}^{x}}$ common on right hand side of the equation we get, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\text{e}}^{x}}\left( {{x}^{2}}+4x+2 \right)$
Therefore, we got the second derivative for the given question.
Therefore,
First derivative of ${{x}^{2}}{{e}^{x}}$ is $\dfrac{dy}{dx}\text{ = }x\left( x+2 \right){{e}^{x}}\text{ }$
Second derivative of ${{x}^{2}}{{e}^{x}}$ is $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\text{e}}^{x}}\left( {{x}^{2}}+4x+2 \right)$

 Note:
We should be well known about the product rule formula and its application in the derivative problems. Similarly to the product rule used in this question we have a division rule given as $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\left( \dfrac{d}{dx}u \right)-u\left( \dfrac{d}{dx}v \right)}{{{v}^{2}}}$ which can be used in questions of this type.