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}} $ .