Search any question & find its solution
Question:
Answered & Verified by Expert
If $\sin x \cos y=\frac{1}{2}$, then what is the value of $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ ?
Options:
Solution:
1994 Upvotes
Verified Answer
The correct answer is:
$-4$
$\sin x \cos y=\frac{1}{2}$
Differentiating both the sides
$\sin x(-\sin y) \frac{d y}{d x}+\cos y \cos x=0$
$\frac{d y}{d x}=\frac{\cos x \cos y}{\sin x \sin y}=\cot x \cot y$
$\frac{d^{2} y}{d x^{2}}=\cot x\left(-\operatorname{cosec}^{2} y\right) \frac{d y}{d x}+\cot y\left(-\operatorname{cosec}^{2} x\right)$
when the point is $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$
then $\mathrm{x}=\frac{\pi}{4}$ and $\mathrm{y}=\frac{\pi}{4}$
$\frac{d y}{d x}=\cot \frac{\pi}{4} \cdot \cot \frac{\pi}{4}=1$
So, $\frac{d^{2} y}{d x^{2}}=\cot \frac{\pi}{4}\left(-\cos e c^{2} \frac{\pi}{4}\right) 1+\cot \frac{\pi}{4}\left(-\cos e c^{2} \frac{\pi}{4}\right)$
$=1(-2)+1 .(-2)=-4$
Differentiating both the sides
$\sin x(-\sin y) \frac{d y}{d x}+\cos y \cos x=0$
$\frac{d y}{d x}=\frac{\cos x \cos y}{\sin x \sin y}=\cot x \cot y$
$\frac{d^{2} y}{d x^{2}}=\cot x\left(-\operatorname{cosec}^{2} y\right) \frac{d y}{d x}+\cot y\left(-\operatorname{cosec}^{2} x\right)$
when the point is $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$
then $\mathrm{x}=\frac{\pi}{4}$ and $\mathrm{y}=\frac{\pi}{4}$
$\frac{d y}{d x}=\cot \frac{\pi}{4} \cdot \cot \frac{\pi}{4}=1$
So, $\frac{d^{2} y}{d x^{2}}=\cot \frac{\pi}{4}\left(-\cos e c^{2} \frac{\pi}{4}\right) 1+\cot \frac{\pi}{4}\left(-\cos e c^{2} \frac{\pi}{4}\right)$
$=1(-2)+1 .(-2)=-4$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.