Question

If $u = \cos^3 x$ and $v = \sin^3 x$, then $\left(\frac{dv}{du}\right)_{x = \frac{\pi}{4}}$ is equal to

• A. $-2$
• B. $2$
• C. $1$
• D. $-1$

Hint:

At $x = \frac{\pi}{4}$, both $\sin x$ and $\cos x$ are equal to $\frac{1}{\sqrt{2}}$. Whenever two parametric functions have a symmetric structure with respect to sine and cosine, their derivative at $x = 45^\circ$ will almost always evaluate simply to $1$ or $-1$. This pattern saves valuable time during exams!

Answer 1

The Correct Option is D

Step 1: Recognise a parametric pair.
Here $u=\cos^3x$ and $v=\sin^3x$ both depend on the parameter $x$, so $\dfrac{dv}{du}=\dfrac{dv/dx}{du/dx}$.
Step 2: Differentiate $v$ using the chain rule.
$\dfrac{dv}{dx}=3\sin^2x\cdot\cos x$.
Step 3: Differentiate $u$.
$\dfrac{du}{dx}=3\cos^2x\cdot(-\sin x)=-3\cos^2x\sin x$.
Step 4: Form the ratio and cancel.
\[ \frac{dv}{du}=\frac{3\sin^2x\cos x}{-3\cos^2x\sin x}=-\frac{\sin x}{\cos x}=-\tan x. \]
Step 5: Evaluate at $x=\dfrac{\pi}{4}$.
$-\tan\dfrac{\pi}{4}=-1$.
Step 6: Conclude.
The required derivative is $-1$. \[ \boxed{\left(\dfrac{dv}{du}\right)_{x=\pi/4}=-1} \]