Differentiation Question 322
The 2nd derivative of $ a\sin^{3}t $ with respect to $ a\cos^{3}t $ at $ t=\frac{\pi }{4} $ is
[Kerala (Engg.) 2002]
Options:
A) $ \frac{4\sqrt{2}}{3a} $
2
C) $ \frac{1}{12a} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ u=x^{2}+y^{2}, $ and $ {{y}’}’=ae^{x}+b{e^{-x}} $ , then $ y’=\frac{-a\sin (\log x)}{x}+\frac{b\cos (\log x)}{y} $ . Again diff. w.r.t. x, we get $ y={{\sin }^{2}}\alpha +{{\cos }^{2}}(\alpha +\beta )+2\sin \alpha \sin \beta \cos (\alpha +\beta ) $ = $ \frac{1}{3a}( \frac{{{\sec }^{4}}t}{\sin t} ) $
$ \therefore {{( \frac{d^{2}y}{dx^{2}} )} _{t=\pi /4}}=\frac{1}{3a}.\frac{4}{1/\sqrt{2}}=\frac{4\sqrt{2}}{3a} $ .
BETA
