Question

How do you find the derivative of $f\left( x \right)={{x}^{3}}{{e}^{x}}$

Answer 1

Hint: We will first recall the concept of differentiation and its various method to solve and then we will find the differentiation of $f\left( x \right)={{x}^{3}}{{e}^{x}}$. To find the differentiation of $f\left( x \right)$ we will first use the multiplication of two different function and then we will use the standard differentiation formulas such as $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$, and $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$.

Complete step-by-step solution:
We will use the concept of methods of differentiation to solve the above question.
Since, we have to find the derivative of $f\left( x \right)={{x}^{3}}{{e}^{x}}$ and we can see that it consists of two different terms one is algebraic which is ${{x}^{3}}$ and other is exponential ${{e}^{x}}$. So, we will apply the multiplication or product rule of differentiation for two different functions.
Let $f\left( x \right)$ and $g\left( x \right)$ be two differentiable function, then differentiation of their product is given as:
$\dfrac{d\left( f\left( x \right).g\left( x \right) \right)}{dx}=f\left( x \right)\times \dfrac{d\left( g\left( x \right) \right)}{dx}+\dfrac{d\left( f\left( x \right) \right)}{dx}\times g\left( x \right)$
So, derivative of $f\left( x \right)={{x}^{3}}{{e}^{x}}$ is equal to:
\[\Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( {{x}^{3}}{{e}^{x}} \right)}{dx}\]
So, we can write $\dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( {{x}^{3}}{{e}^{x}} \right)}{dx}={{x}^{3}}\times \dfrac{d\left( {{e}^{x}} \right)}{dx}+\dfrac{d\left( {{x}^{3}} \right)}{dx}\times {{e}^{x}}$ , by product rule of differentiation.
Now, we know that differentiation of ${{e}^{x}}$ is equal to ${{e}^{x}}$i.e. $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ and from power rule we know that $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$.
So, $\dfrac{d\left( {{x}^{3}} \right)}{dx}=3{{x}^{3-1}}=3{{x}^{2}}$
$\Rightarrow \dfrac{d\left( {{x}^{3}}{{e}^{x}} \right)}{dx}={{x}^{3}}\times \dfrac{d\left( {{e}^{x}} \right)}{dx}+\dfrac{d\left( {{x}^{3}} \right)}{dx}\times {{e}^{x}}$
\[\Rightarrow \dfrac{d\left( {{x}^{3}}{{e}^{x}} \right)}{dx}={{x}^{3}}\times {{e}^{x}}+3{{x}^{2}}\times {{e}^{x}}\]
Now, we will take common ${{e}^{x}}$ from both of the terms, then we will get:
\[\Rightarrow \dfrac{d\left( {{x}^{3}}{{e}^{x}} \right)}{dx}={{e}^{x}}\left( {{x}^{3}}+3{{x}^{2}} \right)\]
Now, we can see that ${{x}^{2}}$ is common to both the terms ${{x}^{3}}$ and $3{{x}^{2}}$, so will take it as common.
\[\Rightarrow \dfrac{d\left( {{x}^{3}}{{e}^{x}} \right)}{dx}={{e}^{x}}{{x}^{2}}\left( x+3 \right)\]
Hence, \[{{e}^{x}}{{x}^{2}}\left( x+3 \right)\] is the required derivative of $f\left( x \right)={{x}^{3}}{{e}^{x}}$.
This is our required solution.

Note: Students are required to memorize all the standard derivative formulas otherwise they will not be able to solve any question. Also, note that the product rule can be extended to more than two functions. Let us say that $f\left( x \right),g\left( x \right)\text{ and }h\left( x \right)$ are three differentiable function, then differentiation of their product is given as:
$\dfrac{d\left( f\left( x \right).g\left( x \right).h\left( x \right) \right)}{dx}=f\left( x \right).g\left( x \right).\dfrac{d\left( h\left( x \right) \right)}{dx}+f\left( x \right).\dfrac{d\left( g\left( x \right) \right)}{dx}.h\left( x \right)+\dfrac{d\left( f\left( x \right) \right)}{dx}.g\left( x \right).h\left( x \right)$ .
Similarly, we can write for more than 3 functions.