Question

How do you find the derivative of $({x^2})({e^x})$ ?

Answer 1

Hint : Use the Product rule of derivative by separating the function into two separate functions as $f'(x)$ and $g'(x)$ , Use the below mentioned formula applying product rule.
Formula:
Product rule,
$\dfrac{d}{{dx}}f(x).g(x) = f'(x).g(x) + f(x).g'(x)$
$\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$\dfrac{d}{{dx}}({e^x}) = {e^x}$

Complete step-by-step answer :
Given a function in x
$= ({x^2})({e^x})$
We have to find the first derivative of the above equation
Let $f(x) = {x^2}$ and $g(x) = {e^x}$ ,
So we have to find the derivative of a product of two functions.
We’ll use product Rule to calculate the derivative of product of function
According to product rule,
$f'(x).g(x) + f(x).g'(x)$ -(1)
For this, first we have to find the $f'(x)$ and $g'(x)$
$f(x) = {x^2}$
Using derivative rule $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$f'(x) = 2x$ -(2)
And
$g(x) = {e^x}$
Using derivative rule $\dfrac{d}{{dx}}({e^x}) = {e^x}$
$g'(x) = {e^x}$ -(3)
Now putting value of $f'(x)$ and $g'(x)$ from equation (2) and (3) in product rule (1)
$= \dfrac{d}{{dx}}({x^2})({e^x}) \\\ = \dfrac{d}{{dx}}f(x).g(x) \\\ = f'(x).g(x) + f(x).g'(x) \\\ = (2x)({e^x}) + ({x^2})({e^x}) \;$
Taking common ${e^x}$ from both of the terms
$= ({e^x})(2x + {x^2})$
Therefore, the derivative of $({x^2})({e^x})$ is equal to $({e^x})(2x + {x^2})$ .
So, the correct answer is “ $({e^x})(2x + {x^2})$ ”.

Note : DifferentiationIt is a method by which we can find the derivative of the function. It is a process through which we can find the instantaneous rate of change in function based on one of its variables.
Let y = f(x) be a function of x. So the rate of change of $y$ per unit change in $x$ is given by:
$\dfrac{{dy}}{{dx}}$ .