Question

The derivative of $\cos^{3} x$ w.r.t $\sin^{3} x$ is
A) $-\cot x$
B) $\cot x$
C) $\tan x$
D) $-\tan x$

Answer 1

Hint: In this question it is given that We have to find the derivative of $\cos^{3} x$ w.r.t $\sin^{3} x$, i.e, $\dfrac{du}{dv} =\dfrac{\left( \dfrac{du}{dx} \right) }{\left( \dfrac{dv}{dx} \right) }$.......(1)
Where u=$\cos^{3} x$ and v=$\sin^{3} x$,
So to find the solution we have to first find the values of $\dfrac{du}{dx} \ and\ \dfrac{dv}{dx}$.
Complete step-by-step solution:
So first of all we are going to find the value of $\dfrac{du}{dx}$.
Therefore,
$\dfrac{du}{dx} =\dfrac{d}{dx} \left( \cos^{3} x\right)$
=$3\cos^{2} x\cdot \dfrac{d}{dx} \left( \cos x\right)$ [ using chain rule]
=$3\cos^{2} x\left( -\sin x\right)$ [ since,$\dfrac{d}{dx} \left( \cos x\right) =-\sin x$]
=$-3\sin x\cos^{2} x$.......(2)
Now,
$\dfrac{dv}{dx} =\dfrac{d}{dx} \left( \sin^{3} x\right)$
=$3\sin^{2} x.\dfrac{d}{dx} \left( \sin x\right)$ [using chain rule]
=$3\sin^{2} x.\cos x$....(3) [since, $\dfrac{d}{dx} \left( \sin x\right) =\cos x$]
Now by putting the values of $\dfrac{du}{dx} \ and\ \dfrac{dv}{dx}$ in equation (1), we get,
$\dfrac{du}{dv} =\dfrac{\left( \dfrac{du}{dx} \right) }{\left( \dfrac{dv}{dx} \right) }$
=$\dfrac{-3\sin x\cos^{2} x}{3\sin^{2} x\cos x}$
=$-\dfrac{\cos x}{\sin x}$
=$-\cot x$ [ since, $\dfrac{\cos x}{\sin x} =\cot x$]
So our required result is $-\cot x$.
Hence the correct option is option A.
Note: While solving you need to know the basic formulas of derivative that we have already mentioned while solving also we have mentioned about chain rule which implies that, if ‘y’ be the function of ‘u’ and ‘u’ be the be the function of ‘x’, then
$\dfrac{dy}{dx} =\dfrac{dy}{du} \cdot \dfrac{du}{dx}$