Question

Find the first two derivatives of $2\sin x\cos x$.

Answer 1

Step 1: Compute the first derivative:

Suppose $y=2\sin \left(x\right)\cos \left(x\right)$

Now, we are using the product rule to find the derivatives.

The formula of the product rule:$uv\text{'}+vu\text{'}$

Here $u=2\sin \left(x\right)$ and $v=\cos \left(x\right)$.

Now, apply the product rule to find the first derivative,

$$\begin{gathered} y\text{'}=2\sin \left(x\right)\times \frac{d}{dx}\left[cos\right(x\left)\right]+\cos \left(x\right)\times \frac{d}{dx}\left[2\sin \right(x\left)\right] \\ y\text{'}=2\sin \left(x\right)\times [-\sin (x\left)\right]+\cos \left(x\right)\times \left[2cos\right(x\left)\right] \end{gathered}$$

Since the formula of $\frac{d}{dx}\left[\cos \right(x\left)\right]=-\sin \left(x\right)$ and $\frac{d}{dx}\left[\sin \right(x\left)\right]=\cos \left(x\right)$.

After simplifying the above expression we get,

$y\text{'}=2[{\cos }^{2}x-{\sin }^{2}x]$

$y\text{'}=2\cos \left(2x\right)\left[\because \cos 2x=[{\cos }^{2}x-{\sin }^{2}x]\right]$

Step 2: Compute the second derivative:

Now, again apply the derivative on the first derivative and we get,

$$\begin{gathered} y"=2\times \frac{d}{dx}\left[\cos \right(2x\left)\right] \\ y"=-2\times \sin 2x\times \frac{d}{dx}\left(2x\right) \\ y"=-4\times \sin \left(2x\right) \end{gathered}$$

Hence, the first two derivative of $y=2\sin \left(x\right)\cos \left(x\right)$ is $y"=-4\sin \left(2x\right)$.