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

Last updated date: 17th Jun 2024
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Hint: In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument, or we can say it is the rate of change of a function. Derivation of two functions multiplied with each other can be executed by simplifying them into a single function using Product rule.

Here we have the function as $y=x{{e}^{x}}$
As we can see the function is a multiplication of two function , that is $x$ and ${{e}^{x}}$
The Product rule of derivative is given as,
$\Rightarrow \dfrac{d(uv)}{dx}=u\dfrac{d(v)}{dx}+v\dfrac{d(u)}{dx}$
Where $u$ and $v$ are two functions/variables and $\dfrac{d}{dx}$ is derivative with respect to $x$.
In the above question the two functions can be $x$ and ${{e}^{x}}$ can be taken as $u$ and $v$.
We will apply the product rule to find the derivative of the above equation.
That is, $y=(x)({{e}^{x}})\Leftrightarrow y=(u)(v)$
Applying product rule we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d(x{{e}^{x}})}{dx}$
$\Rightarrow \dfrac{d(x{{e}^{x}})}{dx}=x\dfrac{d({{e}^{x}})}{dx}+{{e}^{x}}\dfrac{d(x)}{dx}$
Since we know that $\dfrac{d(x)}{dx}=1$ and $\dfrac{d({{e}^{x}})}{dx}={{e}^{x}}$ , derivative of $x$ with respect to $x$ is 1.
And we also know that the exponential function has the special property that its derivative is the function itself.
Substituting the values of derivatives in the above equation we get,
$\Rightarrow \dfrac{d(x{{e}^{x}})}{dx}=x({{e}^{x}})+{{e}^{x}}(1)$
$\Rightarrow \dfrac{d(x{{e}^{x}})}{dx}=x{{e}^{x}}+{{e}^{x}}$
Taking ${{e}^{x}}$ as common
$\Rightarrow \dfrac{d(x{{e}^{x}})}{dx}={{e}^{x}}(x+1)$

Hence the derivative of the function $y=x{{e}^{x}}$ is ${{e}^{x}}(x+1)$.

Note: In product rule the order of function does not matter, that is the values of $u$ and $v$ can be interchanged and we will still get the same answer. For example, in the above question, we took $u$ and $v$ as $x$ and ${{e}^{x}}$ respectively, now interchanging the values , $u={{e}^{x}},v=x$ we get
$\Rightarrow \dfrac{d({{e}^{x}}x)}{dx}$
$\Rightarrow \dfrac{d({{e}^{x}}x)}{dx}={{e}^{x}}\dfrac{d(x)}{dx}+x\dfrac{d({{e}^{x}})}{dx}$
$\Rightarrow \dfrac{d({{e}^{x}}x)}{dx}={{e}^{x}}(1)+x({{e}^{x}})$
Taking ${{e}^{x}}$ as common
$\Rightarrow \dfrac{d({{e}^{x}}x)}{dx}={{e}^{x}}(1+x)$ , which is the same answer as obtained above.