How do you differentiate $g\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)$ using the product rule ?
Answer
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Hint: Differentiation is a process where we find the instantaneous rate of change in function based on one of its variables. It is a method finding the derivative of a function. The most common example is the rate of change of displacement with respect to time, called velocity. If $x$ is a variable and $y$ is another variable ,then the rate of change of $x$ with respect $y$is given by $\dfrac{dy}{dx}$ . For the function given to us, we also have to make use of product rules.
Complete step by step solution:
Since our function is a product of two functions, we need to make use of the product rule of differentiation in order to differentiate it.
As per the product rule, if a function $h\left( x \right)$ is a product of two functions namely $u\left( x \right),v\left( x \right)$ then the derivative of the function is as follows :
$\begin{align}
& \Rightarrow h\left( x \right)=u\left( x \right)v\left( x \right) \\
& \Rightarrow h'\left( x \right)=u'\left( x \right)v\left( x \right)+u\left( x \right)v'\left( x \right) \\
\end{align}$
The function that we are given is $g\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)$.
Upon comparing , our $u\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)$ and our $v\left( x \right)=\left( 2x+2{{x}^{2}} \right)$ .
Before proceeding further, let us look at the derivative of the function ${{e}^{{{x}^{2}}}}$. We know the derivative of ${{e}^{x}}$ .
Let us assume the power of $e$ which is ${{x}^{2}}$ to be $y$. So $y$ is a function of $x$. Now let us differentiate this.
Upon doing so, we get the following :
$\Rightarrow {{e}^{{{x}^{2}}}}={{e}^{y}}$
Let us differentiate ${{e}^{y}}$ with respect to $x$.
$\Rightarrow \dfrac{d\left( {{e}^{y}} \right)}{dx}={{e}^{y}}\dfrac{dy}{dx}$
We have to differentiate $y$ too since it is a function of $x$.
$\begin{align}
& \Rightarrow y={{x}^{2}} \\
& \Rightarrow \dfrac{dy}{dx}=2x \\
\end{align}$
Now let us substitute back the value of $y$ and $\dfrac{dy}{dx}$.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow \dfrac{d\left( {{e}^{y}} \right)}{dx}={{e}^{y}}\dfrac{dy}{dx} \\
& \Rightarrow \dfrac{d\left( {{e}^{{{x}^{2}}}} \right)}{dx}={{e}^{{{x}^{2}}}}2x \\
\end{align}$
Now let us use the product rule and carry out the differentiation of $g\left( x \right)$.
Upon differentiating the function with respect to $x$, we get the following :
\[\begin{align}
& \Rightarrow g\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right) \\
& \Rightarrow g'\left( x \right)=\dfrac{d\left( \left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \right)}{dx}\left( 2x+2{{x}^{2}} \right)+\dfrac{d\left( \left( 2x+2{{x}^{2}} \right) \right)}{dx}\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
& \Rightarrow g'\left( x \right)=\left( 2\left( 2x{{e}^{{{x}^{2}}}} \right)+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)+\left( 2+4x \right)\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
\end{align}\]
Let us multiply this.
Upon doing so, we get the following :
\[\begin{align}
& \Rightarrow g\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right) \\
& \Rightarrow g'\left( x \right)=\dfrac{d\left( \left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \right)}{dx}\left( 2x+2{{x}^{2}} \right)+\dfrac{d\left( \left( 2x+2{{x}^{2}} \right) \right)}{dx}\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
& \Rightarrow g'\left( x \right)=\left( 2\left( 2x{{e}^{{{x}^{2}}}} \right)+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)+\left( 2+4x \right)\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
& \Rightarrow g'\left( x \right)=\left( 4x{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)+\left( 2+4x \right)\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
& \Rightarrow g'\left( x \right)=8{{x}^{2}}{{e}^{{{x}^{2}}}}+8{{x}^{3}}{{e}^{{{x}^{2}}}}+8x{{e}^{x}}+8{{e}^{x}}{{x}^{2}}+4{{e}^{{{x}^{2}}}}+8{{e}^{x}}+8x{{e}^{{{x}^{2}}}}+16x{{e}^{x}} \\
& \Rightarrow g'\left( x \right)=8{{x}^{2}}{{e}^{{{x}^{2}}}}+8{{x}^{3}}{{e}^{{{x}^{2}}}}+24x{{e}^{x}}+8{{e}^{x}}{{x}^{2}}+4{{e}^{{{x}^{2}}}}+8{{e}^{x}}+8x{{e}^{{{x}^{2}}}} \\
\end{align}\]
$\therefore $ Hence, the derivative of $g\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)$ using the product rule is \[g'\left( x \right)=8{{x}^{2}}{{e}^{{{x}^{2}}}}+8{{x}^{3}}{{e}^{{{x}^{2}}}}+24x{{e}^{x}}+8{{e}^{x}}{{x}^{2}}+4{{e}^{{{x}^{2}}}}+8{{e}^{x}}+8x{{e}^{{{x}^{2}}}}\].
Note: Instead of using the product the rule, we can just multiply the terms of one bracket with the terms of another and then just differentiate each term like how we normally do. This can be a little less confusing than the product rule. But we should remember all the rules of differentiation. We should also remember all the derivatives of all the functions to solve a question quickly in the exam. We should also be able to differentiate different kinds of functions. Practice is needed.
Complete step by step solution:
Since our function is a product of two functions, we need to make use of the product rule of differentiation in order to differentiate it.
As per the product rule, if a function $h\left( x \right)$ is a product of two functions namely $u\left( x \right),v\left( x \right)$ then the derivative of the function is as follows :
$\begin{align}
& \Rightarrow h\left( x \right)=u\left( x \right)v\left( x \right) \\
& \Rightarrow h'\left( x \right)=u'\left( x \right)v\left( x \right)+u\left( x \right)v'\left( x \right) \\
\end{align}$
The function that we are given is $g\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)$.
Upon comparing , our $u\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)$ and our $v\left( x \right)=\left( 2x+2{{x}^{2}} \right)$ .
Before proceeding further, let us look at the derivative of the function ${{e}^{{{x}^{2}}}}$. We know the derivative of ${{e}^{x}}$ .
Let us assume the power of $e$ which is ${{x}^{2}}$ to be $y$. So $y$ is a function of $x$. Now let us differentiate this.
Upon doing so, we get the following :
$\Rightarrow {{e}^{{{x}^{2}}}}={{e}^{y}}$
Let us differentiate ${{e}^{y}}$ with respect to $x$.
$\Rightarrow \dfrac{d\left( {{e}^{y}} \right)}{dx}={{e}^{y}}\dfrac{dy}{dx}$
We have to differentiate $y$ too since it is a function of $x$.
$\begin{align}
& \Rightarrow y={{x}^{2}} \\
& \Rightarrow \dfrac{dy}{dx}=2x \\
\end{align}$
Now let us substitute back the value of $y$ and $\dfrac{dy}{dx}$.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow \dfrac{d\left( {{e}^{y}} \right)}{dx}={{e}^{y}}\dfrac{dy}{dx} \\
& \Rightarrow \dfrac{d\left( {{e}^{{{x}^{2}}}} \right)}{dx}={{e}^{{{x}^{2}}}}2x \\
\end{align}$
Now let us use the product rule and carry out the differentiation of $g\left( x \right)$.
Upon differentiating the function with respect to $x$, we get the following :
\[\begin{align}
& \Rightarrow g\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right) \\
& \Rightarrow g'\left( x \right)=\dfrac{d\left( \left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \right)}{dx}\left( 2x+2{{x}^{2}} \right)+\dfrac{d\left( \left( 2x+2{{x}^{2}} \right) \right)}{dx}\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
& \Rightarrow g'\left( x \right)=\left( 2\left( 2x{{e}^{{{x}^{2}}}} \right)+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)+\left( 2+4x \right)\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
\end{align}\]
Let us multiply this.
Upon doing so, we get the following :
\[\begin{align}
& \Rightarrow g\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right) \\
& \Rightarrow g'\left( x \right)=\dfrac{d\left( \left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \right)}{dx}\left( 2x+2{{x}^{2}} \right)+\dfrac{d\left( \left( 2x+2{{x}^{2}} \right) \right)}{dx}\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
& \Rightarrow g'\left( x \right)=\left( 2\left( 2x{{e}^{{{x}^{2}}}} \right)+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)+\left( 2+4x \right)\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
& \Rightarrow g'\left( x \right)=\left( 4x{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)+\left( 2+4x \right)\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right) \\
& \Rightarrow g'\left( x \right)=8{{x}^{2}}{{e}^{{{x}^{2}}}}+8{{x}^{3}}{{e}^{{{x}^{2}}}}+8x{{e}^{x}}+8{{e}^{x}}{{x}^{2}}+4{{e}^{{{x}^{2}}}}+8{{e}^{x}}+8x{{e}^{{{x}^{2}}}}+16x{{e}^{x}} \\
& \Rightarrow g'\left( x \right)=8{{x}^{2}}{{e}^{{{x}^{2}}}}+8{{x}^{3}}{{e}^{{{x}^{2}}}}+24x{{e}^{x}}+8{{e}^{x}}{{x}^{2}}+4{{e}^{{{x}^{2}}}}+8{{e}^{x}}+8x{{e}^{{{x}^{2}}}} \\
\end{align}\]
$\therefore $ Hence, the derivative of $g\left( x \right)=\left( 2{{e}^{{{x}^{2}}}}+4{{e}^{x}} \right)\left( 2x+2{{x}^{2}} \right)$ using the product rule is \[g'\left( x \right)=8{{x}^{2}}{{e}^{{{x}^{2}}}}+8{{x}^{3}}{{e}^{{{x}^{2}}}}+24x{{e}^{x}}+8{{e}^{x}}{{x}^{2}}+4{{e}^{{{x}^{2}}}}+8{{e}^{x}}+8x{{e}^{{{x}^{2}}}}\].
Note: Instead of using the product the rule, we can just multiply the terms of one bracket with the terms of another and then just differentiate each term like how we normally do. This can be a little less confusing than the product rule. But we should remember all the rules of differentiation. We should also remember all the derivatives of all the functions to solve a question quickly in the exam. We should also be able to differentiate different kinds of functions. Practice is needed.
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