How do you find the derivative \[y=2x\cos x\]?

Hint: To find the derivative of above equation \[y=2x\cos x\] with respect to $x$. The derivative of the function is defined as if we have $x$ and $y$ are real numbers then if the graph of f is plotted with respect to the $x$ then the slope of this graph at each point is known as derivative of function.we will use product Rule of differentiation .let`s suppose $f\left(x \right)$ is equal to $2x$ and $g\left(x \right)$ is equal to $\cos x$. So whenever if we have two functions like \[f\left (x \right)\] and\ [g\left(x \right)\] and both are in multiplication form like$f\left(x \right)\cdot g\left(x \right)$ then the derivative of these type of function is as:
 \[\dfrac{d\left( f\left( x \right)g\left( x \right) \right)}{dx}=f\grave{\ }\left( x \right)g\left( x \right)+f\left( x \right)g\grave{\ }\left( x \right)\] .where $f\grave{\ }\left( x \right)$ is the differentiation of $2x$ and $g\grave{\ }\left( x \right)$ is the differentiation of $\cos \left( x \right)$ and $f\left( x \right)=2x$ and $g\left( x \right)=\cos \left( x \right)$.

Complete step by step solution:
Here the given equation is:
$y=2x\cos x$
In the given equation $x$ is an independent variable and $y$ is the dependent variable. The value of $y$ changes as $x$ changes.
Now by applying the product rule\[\dfrac{d\left( f\left( x \right)g\left( x \right) \right)}{dx}=f\grave{\ }\left( x \right)g\left( x \right)+f\left( x \right)g\grave{\ }\left( x \right)\] on $f\left( x \right)=2x$ and$g\left( x \right)=\cos \left( x \right)$ , we get:
    \[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\left( 2x \right)\grave{\ }\left( \cos x \right)+\left( 2x \right)\left( \cos x \right)\grave{\ } \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( 2x \right)}{dx}\left( \cos x \right)+\left( 2x \right)\dfrac{d\left( \cos x \right)}{dx}..........\left( 1 \right) \\
\end{align}\]
Now by solving $\dfrac {d\left (2x \right)}{dx}$ and $\dfrac {d\left (\cos x \right)}{dx}$ we get:
\[\Rightarrow \dfrac{d\left (2x \right)}{dx}=2\] and $\dfrac {d\left (\cos x \right)}{dx}=-\sin x$
Putting these values of differentiation in equation (1), we get,
$\Rightarrow \dfrac{dy}{dx}=2\cos x-2x\sin x$
Hence by using the product rule of differentiation, we get the derivative of \[y=2x\cos x\] is $2\cos x-2x\sin x$ .

Note:
In order to find the derivative of an equation we can go wrong by writing the incorrect differentiation of given functions. Like Sometimes we forget to write the minus sign in the differentiation of \[\cos x\]. By using product rule \[\dfrac{d\left( f\left( x \right)g\left( x \right) \right)}{dx}=f\grave{\ }\left( x \right)g\left( x \right)+f\left( x \right)g\grave{\ }\left( x \right)\]
Must remember positive signs between the above equations, sometimes we make mistakes in this by taking negative signs. And we should know the differentiation of every function. Then finding the derivative of any equation became easy.