Differentiation Question 407
Question: Let $ y=t^{10}+1 $ and $ x=t^{8}+1, $ then $ \frac{d^{2}y}{dx^{2}} $ is
[UPSEAT 2004]
Options:
A) $ \frac{5}{2}t $
B) $ 20t^{8} $
C) $ \frac{5}{16t^{6}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Here $ y=t^{10}+1 $ and $ x=t^{8}+1 $
$ \therefore $ $ t^{8}=x-1 $
$ \Rightarrow t^{2}={{(x-1)}^{1/4}} $
So, $ y={{(x-1)}^{5/4}}+1 $
Differentiate both sides w.r.t. x, $ \frac{dy}{dx}=\frac{5}{4}{{(x-1)}^{1/4}} $
Again, differentiate both sides w.r.t. x, $ \frac{d^{2}y}{dx^{2}}=\frac{5}{16}{{(x-1)}^{-3/4}} $
$ \frac{d^{2}y}{dx^{2}}=\frac{5}{16{{(x-1)}^{3/4}}}=\frac{5}{16{{(t^{2})}^{3}}}=\frac{5}{16t^{6}}. $
BETA
