Question

How do you differentiate $f(x)=x{{e}^{-2x}}$ using the product rule?

Answer 1

To solve this problem student should be familiar with the product rule for differentiation, which is $\dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=f\left( x \right).\dfrac{d}{dx}\left( g\left( x \right) \right)+\dfrac{d}{dx}\left( f\left( x \right) \right)\cdot g\left( x \right)$ . Then application of this formula for the given problem with considering x as first term and ${{e}^{-2x}}$ as the second term. Then we can simplify this to get the final answer.

Complete step by step solution:
The two terms given in the question are x and ${{e}^{-2x}}$.
The product rule states that the derivative of the product of the two will equal the following:
We know the general form of product rule which is as below,
$\dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)\cdot g\left( x \right)+f\left( x \right).\dfrac{d}{dx}\left( g\left( x \right) \right)$
So for this question we can differentiate the given functions using product rule as shown,
$f'(x)=\dfrac{d}{dx}\left( x \right)\cdot {{e}^{-2x}}+x\cdot \dfrac{d}{dx}\left( {{e}^{-2x}} \right)$
Let's take a moment to calculate both of the derivatives inside of our product rule statement. Initially let us consider our first function which is x,
$\dfrac{d}{dx}\left( x \right)=1$
Now, the next derivative will require the chain rule. Also, remember that the derivative of ${{e}^{x}}$is ${{e}^{x}}$. So derivative of second function,
We know from chain rule, $g\left( h\left( x \right) \right)=g'\left( h\left( x \right) \right).h'\left( x \right)$
$\dfrac{d}{dx}\left[ {{e}^{-2x}} \right]={{e}^{-2x}}.\dfrac{d}{dx}\left[ -2x \right]=-2\left( {{e}^{-2x}} \right)$
Now that we know what both our derivatives are equal to, let's plug them back into our equation.
$f'(x)=1.{{e}^{-2x}}+x.-2x\left( {{e}^{-2x}} \right)$
Let us now simplify the above substitution to get,
$\Rightarrow$ $f'(x)={{e}^{-2x}}-2{{x}^{2}}\left( {{e}^{-2x}} \right)$
$\Rightarrow$ $f'(x)={{e}^{-2x}}\left( 1-2{{x}^{2}} \right)$
$\Rightarrow$ $f'(x)=\dfrac{\left( 1-2{{x}^{2}} \right)}{{{e}^{2x}}}$
So, the differentiation of the function $f(x)=x{{e}^{-2x}}$ using the product rule is $\dfrac{\left( 1-2{{x}^{2}} \right)}{{{e}^{2x}}}$ .

Note: Students should be aware to use the product rule to get the solution for this problem. In this type of problem one should be well versed with the differentiation of the basic functions like x and ${{e}^{-2x}}$ .We also have quotient rule similar to that of product rule where it is given by $\dfrac{d}{dx}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{g\left( x \right)\dfrac{d}{dx}\left( f\left( x \right) \right)-f\left( x \right).\dfrac{d}{dx}\left( g\left( x \right) \right)}{{{\left( g\left( x \right) \right)}^{2}}}$ .