Question

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

Answer 1

In the above problem, we are asked to find the derivative of $y={{x}^{2}}{{e}^{-x}}$. For that, we are going to differentiate on both the sides with respect to x. And the derivative on the R.H.S of the above equation is done by using the product rule because on the R.H.S of the above equation two functions $\left( {{x}^{2}},{{e}^{-x}} \right)$ are written in the product form. We know that the application of product rule in two functions say (f(x) & g(x)) is equal to: $\dfrac{d\left( f\left( x \right)g\left( x \right) \right)}{dx}=f\left( x \right)g'\left( x \right)+g\left( x \right)f'\left( x \right)$.

Complete step by step answer:
The function that we have to take derivative of is given as follows:
$y={{x}^{2}}{{e}^{-x}}$
Now, differentiating both the sides with respect to x we get,
$\dfrac{dy}{dx}=\dfrac{d\left( {{x}^{2}}{{e}^{-x}} \right)}{dx}$
As you can see that on R.H.S of the above equation, two functions $\left( {{x}^{2}},{{e}^{-x}} \right)$ are written in the product form so we are going to apply product rule to derive the R.H.S.
We know that product rule between two functions (f(x) & g(x)) is equal to:
$\dfrac{d\left( f\left( x \right)g\left( x \right) \right)}{dx}=f\left( x \right)g'\left( x \right)+g\left( x \right)f'\left( x \right)$
Now, using the above product rule to derivate ${{x}^{2}}{{e}^{-x}}$ with respect to x we get,
On comparing ${{x}^{2}}{{e}^{-x}}$ with the above product rule $\left( f\left( x \right)={{x}^{2}},g\left( x \right)={{e}^{-x}} \right)$.
$\dfrac{dy}{dx}={{x}^{2}}\left( {{e}^{-x}} \right)'+{{e}^{-x}}\left( {{x}^{2}} \right)'$
$\Rightarrow \dfrac{dy}{dx}={{x}^{2}}\left( -{{e}^{-x}} \right)+{{e}^{-x}}\left( 2x \right)$
Taking $x{{e}^{-x}}$ as common from R.H.S of the above equation we get,
$\dfrac{dy}{dx}=x{{e}^{-x}}\left( -x+2 \right)$
From the above differentiation, we have taken the derivative of given equation and is equal to:
$\dfrac{dy}{dx}=x{{e}^{-x}}\left( -x+2 \right)$

Note:
In the haste of solving the above problem, you might mistakenly interpret the derivative as “integration” and might do the integration of the above problem so make sure you won’t make this mistake in the examination.
Also, while differentiating ${{e}^{-x}}$ with respect to x, you might forget to write the negative sign in the differentiation. Like, you might have done the differentiation as follows:
$\dfrac{d{{e}^{-x}}}{dx}={{e}^{-x}}$
This is the wrong derivative of ${{e}^{-x}}$ with respect to x so keep in mind this plausible mistake and be alert not to repeat in the examination.