
How do you find the derivative of $2x\cos x$?
Answer
493.8k+ views
Hint: In this problem we need to calculate the derivative of the given function. We can observe that the given function has two different functions in multiplication. The first one is the algebraic function which is $2x$. The second one is the trigonometric function which is $\cos x$. We will write the both functions as two separate variables. Now we will differentiate the both the variables with respect to $x$ by using the differentiation formulas. After calculating the derivatives of each function, we will use the $uv$ formula of the differentiation which is $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. We will substitute the differentiation values of each variable in the above equation and simplify them to get the required result.
Complete step-by-step solution:
Given that, $2x\cos x$.
In the above function we have two functions, the first one is $2x$ second one is $\cos x$.
Let us assume that
$u=2x$, $v=\cos x$
Differentiating the above equations with respect to $x$, then we will get
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( 2x \right)$, $\Rightarrow \dfrac{dv}{dx}=\dfrac{d}{dx}\left( \cos x \right)$
We have the differentiation formulas $\dfrac{d}{dx}\left( ax \right)=a$, $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$. Applying the above formulas in the above equation, then we will get
$\Rightarrow \dfrac{du}{dx}=2$, $\Rightarrow \dfrac{dv}{dx}=-\sin x$
Now differentiating the given function $2x\cos x$ with respect to $x$, then we will have $\dfrac{d}{dx}\left( 2x\cos x \right)$.
Applying the formula $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ in the above value, then we will get
$\Rightarrow \dfrac{d}{dx}\left( 2x\cos x \right)=2x\dfrac{dv}{dx}+\cos x\dfrac{du}{dx}$
Substituting the values, we have, in the above equation, then we will get
$\begin{align}
& \Rightarrow \dfrac{d}{dx}\left( 2x\cos x \right)=2x\left( -\sin x \right)+\cos x\left( 2 \right) \\
& \Rightarrow \dfrac{d}{dx}\left( 2x\cos x \right)=2\cos x-2x\sin x \\
\end{align}$
Hence the derivative of the given value $2x\cos x$ is $2\cos x-2x\sin x$.
Note: In this problem we have calculated the values of $\dfrac{du}{dx}$, $\dfrac{dv}{dx}$ separately which is not necessary for this particular problem. But it is a good practice to calculate them separately. It will be very useful when we have complex functions.
Complete step-by-step solution:
Given that, $2x\cos x$.
In the above function we have two functions, the first one is $2x$ second one is $\cos x$.
Let us assume that
$u=2x$, $v=\cos x$
Differentiating the above equations with respect to $x$, then we will get
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( 2x \right)$, $\Rightarrow \dfrac{dv}{dx}=\dfrac{d}{dx}\left( \cos x \right)$
We have the differentiation formulas $\dfrac{d}{dx}\left( ax \right)=a$, $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$. Applying the above formulas in the above equation, then we will get
$\Rightarrow \dfrac{du}{dx}=2$, $\Rightarrow \dfrac{dv}{dx}=-\sin x$
Now differentiating the given function $2x\cos x$ with respect to $x$, then we will have $\dfrac{d}{dx}\left( 2x\cos x \right)$.
Applying the formula $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ in the above value, then we will get
$\Rightarrow \dfrac{d}{dx}\left( 2x\cos x \right)=2x\dfrac{dv}{dx}+\cos x\dfrac{du}{dx}$
Substituting the values, we have, in the above equation, then we will get
$\begin{align}
& \Rightarrow \dfrac{d}{dx}\left( 2x\cos x \right)=2x\left( -\sin x \right)+\cos x\left( 2 \right) \\
& \Rightarrow \dfrac{d}{dx}\left( 2x\cos x \right)=2\cos x-2x\sin x \\
\end{align}$
Hence the derivative of the given value $2x\cos x$ is $2\cos x-2x\sin x$.
Note: In this problem we have calculated the values of $\dfrac{du}{dx}$, $\dfrac{dv}{dx}$ separately which is not necessary for this particular problem. But it is a good practice to calculate them separately. It will be very useful when we have complex functions.
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