Question

What is the derivative of ${\sin }^{2}x$ with respect of ${\cos }^{2}x$?

• A. $\tan x$
• B. ${\tan }^{2}x$
• C. $-\tan x$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation of correct answer :

Finding derivative of ${\sin }^{2}x$ with respect to ${\cos }^{2}x$ :

Let , $u={\sin }^{2}x$

Now finding, $\frac{du}{dx}=\frac{d\left({\sin }^{2}x\right)}{dx}$

$$\begin{gathered} \Rightarrow \frac{du}{dx}=2\sin x\left(\cos x\right)\left(\text{Using chain rule}\right) \\ =2\sin x\cos x \end{gathered}$$

Let, $v={\cos }^{2}x$

Now finding, $\frac{dv}{dx}=\frac{d\left({\cos }^{2}x\right)}{dx}$

$$\begin{gathered} \Rightarrow \frac{dv}{dx}=2\cos x\left(-\sin x\right)\left(\text{Using chain rule}\right) \\ =-2\sin x\cos x \end{gathered}$$

We need, $\frac{du}{dv}$

Which can be written as

$\frac{du}{dv}=\frac{\left({\displaystyle \frac{du}{dx}}\right)}{\left({\displaystyle \frac{dv}{dx}}\right)}$

So,

$\Rightarrow \frac{du}{dv}=\frac{2\sin x\cos x}{-2\sin x\cos x}=-1$

Hence, the derivative of ${\sin }^{2}x$ with respect to ${\cos }^{2}x$ is $-1$.

Thus, the correct answer is Option (D).