Limits Continuity And Differentiability Question 111
Question: Let $ y=t^{10}+1 $ and $ x=t^{8}+1 $ , then $ \frac{d^{2}y}{dx^{2}} $ is equal to:
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:
We have, $ y=t^{10}+1\therefore \frac{dy}{dt}=10t^{9} $ and $ x=t^{8}+1\Rightarrow \frac{dx}{dt}=8t^{7} $
$ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{10t^{9}}{8t^{7}}=\frac{5t^{10}}{4t^{8}}=\frac{5(y-1)}{4(x-1)}…(i) $ Differentiate (i) with respect to x, we get
$ \therefore \frac{d^{2}y}{dx^{2}}=\frac{5}{4}.\frac{(x-1).\frac{dy}{dx}-(y-1)}{{{(x-1)}^{2}}} $
$ =\frac{5}{4}.\frac{1}{(x-1)}[ \frac{dy}{dx}-\frac{(y-1)}{(x-1)} ] $
$ =\frac{5}{4}.\frac{1}{(x-1)}[ \frac{5}{4}.\frac{(y-1)}{(x-1)}-\frac{(y-1)}{(x-1)} ] $ [Putting (i)] $ =\frac{5}{4}.\frac{(y-1)}{{{(x-1)}^{2}}}( \frac{5}{4}-1 )=\frac{5}{16}.\frac{t^{10}}{{{(t^{8})}^{2}}}=\frac{5}{16t^{6}} $
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