How do you differentiate \[f\left( x \right)=2\sin x\cos x\]?

Hint: Assume \[f\left( x \right)\] as the product of two trigonometric functions \[\sin x\] and \[\cos x\] and consider them as ‘u’ and ‘v’ respectively. Now, apply the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] to get the answer. Use the formula: - \[\dfrac{d\left( \cos x \right)}{dx}=-\sin x\] and \[\dfrac{d\left( \sin x \right)}{dx}=\cos x\]. Use the relation: - \[{{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x\].

Complete step by step answer:
Here, we have been provided with the function \[f\left( x \right)=2\sin x\cos x\] and we are asked to differentiate it.
Now, we can assume the function given as the product of two trigonometric functions i.e., \[\sin x\] and \[\cos x\], which is multiplied to a constant 2. So, we have,
\[\Rightarrow f\left( x \right)=2\times \cos x\times \sin x=2\times \sin x\times \cos x\]
Let us assume \[\sin x\] and \[\cos x\] as ‘u’ and ‘v’ respectively.
So, we have,
\[\Rightarrow f\left( x \right)=2\times u\times v\]
Differentiating both the sides with respect to x, we get,
\[\Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=\dfrac{d\left( 2\times u\times v \right)}{dx}\]
Since, 2 is a constant so it can be taken out of the derivative. Therefore, we have,
\[\Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=2\times \dfrac{d\left( u\times v \right)}{dx}\]
Now, applying the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\], we get,
\[\Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=2\times \left[ u\dfrac{dv}{dx}+v\dfrac{du}{dx} \right]\]
Substituting the assumed values of u and v, we get,
\[\Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=2\times \left[ \sin x\dfrac{d\left( \cos x \right)}{dx}+\cos x\dfrac{d\left( \sin x \right)}{dx} \right]\]
Using the basic formulas: - \[\dfrac{d\left( \cos x \right)}{dx}=-\sin x\] and \[\dfrac{d\left( \sin x \right)}{dx}=\cos x\], we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=2\times \left[ \sin x\left( -\sin x \right)+\cos x\left( \cos x \right) \right] \\
 & \Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=2\times \left[ -{{\sin }^{2}}x+{{\cos }^{2}}x \right] \\
 & \Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=2\times \left[ {{\cos }^{2}}x-{{\sin }^{2}}x \right] \\
\end{align}\]
Using the trigonometric identity: - \[{{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x\], we get,
\[\Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=2\cos 2x\]
Hence, the above relation is our answer.

Note: One may note that we can also solve the above question with an easier method. What we can do is we will use the trigonometric identity \[2\sin x\cos x=\sin 2x\] at the initial stage of the solution. Now, we will use the formula: - \[\dfrac{d\sin \left( ax+b \right)}{dx}=a\cos \left( ax+b \right)\], where ‘a’ is the coefficient of x and ‘b’ is the constant term, to get the answer. Here, a = 2, b = 0. You must remember all the basic formulas of differentiation like chain rule, product rule, \[\dfrac{u}{v}\] etc. rule as these are used everywhere in the topic of calculus.