Question

How do you find the derivative of $y=2x{{e}^{x}}-2{{e}^{x}}$?

Answer 1

To solve the given question, we must know the derivatives of some functions and a rule of derivative. The function whose derivative we should know are $x\And {{e}^{x}}$, their derivative with respect to x are $1\And {{e}^{x}}$ respectively. We should also know the product rule of the differentiation which states that $\dfrac{d\left( f(x)g(x) \right)}{dx}=\dfrac{d\left( f(x) \right)}{dx}g(x)+f(x)\dfrac{d\left( g(x) \right)}{dx}$. We will use these to find the derivative of the given expression.

Complete step-by-step answer:
We are given the expression $y=2x{{e}^{x}}-2{{e}^{x}}$, we need to find its derivative. The given expression has two terms, as we can see both of the terms have $2{{e}^{x}}$ common to them. So, we can take this factor common to both of them, and write the given expression as,
$y=2{{e}^{x}}\left( x-1 \right)$
The given expression is of the form $f(x)g(x)$, that is a product of two functions. We know the product rule states that $\dfrac{d\left( f(x)g(x) \right)}{dx}=\dfrac{d\left( f(x) \right)}{dx}g(x)+f(x)\dfrac{d\left( g(x) \right)}{dx}$. Here we have $f(x)=2{{e}^{x}}\And g(x)=x-1$
To find the $\dfrac{dy}{dx}$ or $\dfrac{d\left( 2{{e}^{x}}\left( x-1 \right) \right)}{dx}$, we need to find $\dfrac{d\left( 2{{e}^{x}} \right)}{dx}$, and $\dfrac{d\left( x \right)}{dx}$.
We know the derivative of ${{e}^{x}}$ with respect to x, is ${{e}^{x}}$ itself. So, $\dfrac{d\left( 2{{e}^{x}} \right)}{dx}=2{{e}^{x}}$. Also, the derivative of x with respect to x is 1, $\dfrac{d\left( x-1 \right)}{dx}=1$.
$\dfrac{dy}{dx}=\dfrac{d\left( 2{{e}^{x}}\left( x-1 \right) \right)}{dx}$
Using the product, we get
$\Rightarrow \dfrac{d\left( 2{{e}^{x}} \right)}{dx}\left( x-1 \right)+2{{e}^{x}}\dfrac{d\left( x-1 \right)}{dx}$
Substituting the values of the derivatives, we get
$\Rightarrow 2{{e}^{x}}\left( x-1 \right)+2{{e}^{x}}\times 1=2x{{e}^{x}}$
Hence, the derivative of the given expression is $2x{{e}^{x}}$.

Note: Here we expressed the given expression in the form ${{e}^{x}}f(x)$. For the expressions of these forms, we can use a trick for differentiation.
$\dfrac{d\left( {{e}^{x}}f(x) \right)}{dx}={{e}^{x}}\left( f(x)+\dfrac{d\left( f(x) \right)}{dx} \right)$.
By using this property, we can find the derivative of these types of expression easily.