Question

How do you differentiate $g\left( x \right)=x{{e}^{2x}}$ using the product rule?

Answer 1

We are given a function $g\left( x \right)=x{{e}^{2x}}$ and we are asked to find the derivative of it using the product rule. To do the solution we will learn about the product rule. Once we have it we will also need the knowledge that how we differentiate x and ${{e}^{2x}}.$ We know that the derivative of x is 1 and derivative of ${{e}^{x}}$ is ${{e}^{x}}.$ Along with this, we will need the knowledge of the chain rule of derivative to get to the final product.

Complete step-by-step solution:
We are given a function given as $g\left( x \right)=x{{e}^{2x}}$ and we are asked to find the derivative of it. We have to use the product rule to find the derivative. Before we move forward we will learn the product rule. The product rule is for those functions in which the single function is formed by the product of 2 functions. We can see that our function is formed by the product of x and ${{e}^{2x}}.$ So, that’s why we need the product rule to find the derivative. So, the product rule says that for the term as u.v, its derivative is given as
$\dfrac{d\left( u.v \right)}{dx}=\dfrac{vd\left( u \right)}{dx}+\dfrac{ud\left( v \right)}{dx}$
So, to find the derivative of $x{{e}^{2x}}$ we consider x as u and ${{e}^{2x}}$ as v.
Now, as we have u = x and $v={{e}^{2x}}$ so we will use the product rule which says
$\dfrac{d\left( u.v \right)}{dx}=\dfrac{vd\left( u \right)}{dx}+\dfrac{ud\left( v \right)}{dx}$
So, we will get,
$\Rightarrow \dfrac{d\left( x{{e}^{2x}} \right)}{dx}={{e}^{2x}}\dfrac{dx}{dx}+x\dfrac{d\left( {{e}^{2x}} \right)}{dx}......\left( i \right)$
Now, we know that $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}},$ so we will get,
$\dfrac{d\left( {{x}^{1}} \right)}{dx}=1......\left( ii \right)$
Also, we will find the derivative of ${{e}^{2x}}.$ We have to use the chain rule as we can see that ${{e}^{2x}}$ is formed by the composition of 2 functions ${{e}^{2x}}$ and 2x.
${{e}^{2x}}={{e}^{x}}.\left( 2x \right)$
For such functions, to differentiate them we use chain rule which is given as $\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)$
So, we consider $f\left( x \right)={{e}^{x}}$ and $g\left( x \right)=2x.$ So, we get, $\dfrac{d\left( {{e}^{2x}} \right)}{dx}={{e}^{2x}}.\left( 2 \right)$ as $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ and $\dfrac{d\left( 2x \right)}{dx}=2.$
So,
$\dfrac{d\left( {{e}^{2x}} \right)}{dx}=2{{e}^{2x}}......\left( iii \right)$
Using (ii) and (iii) in (i), we get,
$\Rightarrow \dfrac{d\left( x{{e}^{2x}} \right)}{dx}={{e}^{2x}}.1+1.\left( 2{{e}^{2x}} \right)$
Simplifying, we get,
$\Rightarrow \dfrac{d\left( x{{e}^{2x}} \right)}{dx}={{e}^{2x}}+2{{e}^{2x}}$
$\Rightarrow \dfrac{d\left( x{{e}^{2x}} \right)}{dx}=3{{e}^{2x}}$
So, we get the derivative of $x{{e}^{2x}}$ as $3{{e}^{2x}}.$

Note: While simplifying the fraction remember we can add like terms only. While calculating derivative we need to be careful while picking the function as if our function is comprised of more than one function then its derivative will be different than usual like $\dfrac{d\left( {{e}^{2x}} \right)}{dx}\ne {{e}^{2x}}$ as it was made up by the composition of ${{e}^{x}}$ and 2x. Also, while differentiating we need to carefully use $d\left( {{x}^{n}} \right)=n.{{x}^{n-1}}.$ Simple error may happen while calculating.