Chapter 2.5, Problem 42E

To determine

To calculate: The derivative of $\sin 2x$

Answer to Problem 42E

The derivative of $\sin 2x$ is $2\cos 2x$

Explanation of Solution

Given information:

The identities are $\sin 2x=2\sin x\cos x$ , $\cos 2x={\cos }^{2}x-{\sin }^{2}x$

Concept used:

The formulae used are $\frac{d}{dx}\left(\sin x\right)=\cos x$ , $\frac{d}{dx}\left(\cos x\right)=-\sin x$

Calculation:

Differentiate the function with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(\sin 2x\right)=\frac{d}{dx}\left(2\sin x\cos x\right)\\ =2\frac{d}{dx}\left(\sin x\cos x\right)\\ =2\left[\sin x\frac{d}{dx}\cos x+\cos x\frac{d}{dx}\sin x\right]\\ =2\left[\sin x\left(-\sin x\right)+\cos x\left(\cos x\right)\right]\end{array}$

Solve further by substituting ${\cos }^{2}x-{\sin }^{2}x$ for $\sin 2x$ .

  $\begin{array}{c}\frac{d}{dx}\left(\sin 2x\right)=2\left({\cos }^{2}x-{\sin }^{2}x\right)\\ =2\cos 2x\end{array}$

Conclusion: The derivative of $\sin 2x$ is $2\cos 2x$