Question

What is the derivative of ${\cos }^{3}x$ with respect to ${\sin }^{3}x$?

• A. $-cotx$
• B. $cotx$
• C. $\tan x$
• D. $-\tan x$

Answer 1

The correct option is A

$-cotx$

Explanation of correct answer :

Finding the derivative of ${\cos }^{3}x$ with respect to ${\sin }^{3}x$ :

Let $u$be a function of $x$ given as $u\left(x\right)={\cos }^{3}x$

and, $v$ be a function of $x$ given as $v\left(x\right)={\sin }^{3}x$

Now finding, $\frac{d}{dx}\left(u\right)=\frac{du}{dx}=\frac{d\left({\cos }^{3}x\right)}{dx}$

$$\begin{gathered} \Rightarrow \frac{du}{dx}=3\left({\cos }^{2}x\right)\times \left(-\sin x\right)\left(\text{Using chain rule}\right) \\ \Rightarrow \frac{du}{dx}=-3{\cos }^{2}x\sin x \end{gathered}$$

Similarly, $\frac{d}{dx}\left(v\right)=\frac{dv}{dx}=\frac{d\left({\sin }^{3}x\right)}{dx}$

$$\begin{gathered} \Rightarrow \frac{dv}{dx}=3\left({\sin }^{2}x\right)\left(\cos x\right)\left(\text{Using chain rule}\right) \\ \Rightarrow \frac{dv}{dx}=3{\sin }^{2}x\cos x \end{gathered}$$

As per question , we need $\frac{d\left({\cos }^{3}x\right)}{d\left({\sin }^{3}x\right)}=\frac{du}{dv}$

$$\begin{gathered} \Rightarrow \frac{du}{dv}=\frac{\left({\displaystyle \frac{du}{dx}}\right)}{\left({\displaystyle \frac{dv}{dx}}\right)} \\ =\frac{-3{\cos }^{2}x\sin x}{3{\sin }^{2}x\cos x} \\ =\frac{-\cos x}{\sin x} \\ =-cotx \end{gathered}$$

Hence, the derivative of ${\cos }^{3}x$ with respect to ${\sin }^{3}x$ is $-cotx$.

Hence, the correct answer is Option(A).