The 2nd derivative of $a \sin ^3 t$ with respect to $a \cos ^3 t$ at $t=\pi / 4$ is

(A) $\frac{4 \sqrt{2}}{3 a}$

(B) 2

(C) $\frac{1}{12 a}$

(D) None of these

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Best answer

**SOLUTION —**

Let $u=a \sin ^3 t$ and $v=a \cos ^3 t$

$\therefore \quad \frac{d u}{d t}=3 a \sin ^2 t \cos t$

and $\frac{d v}{d t}=3 a \cos ^2 t(-\sin t)$

$\Rightarrow \quad \frac{d^2 u}{d t^2}=3 a\left[2 \sin t \cos ^2 t-\sin ^3 t\right]$

and $\frac{d^2 v}{d t^2}=-3 a\left[-2 \cos t \sin ^2 t+\cos ^3 t\right]$

$\therefore \quad \frac{d^2 u}{d^2 v}=\frac{\left[2 \sin t \cos ^2 t-\sin ^3 t\right]}{\left[2 \cos t \sin ^2 t-\cos ^3 t\right]}$

At $t=\pi / 4$,

$\frac{d^2 u}{d^2 v}=\frac{\left[2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2}-\frac{1}{2 \sqrt{2}}\right]}{\left[2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2}-\frac{1}{2 \sqrt{2}}\right]}=1$

So, The correct option will be **(D).**

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