Question

If $y=t^{10}+1$ and $x=t^8+1$, then $\frac{d^{2}y}{d{x}^2}$ is equal to

• A. $\frac{5}{2}t$
• B. $20t^8$
• C. $\frac{5}{16t^6}$
• D. None of these

Answer 1

Answer: (C)

$\frac{5}{16t^6}$

Here, $y=t^{10}+1$ and $x=t^8+1$
$\therefore t^8=x-1\Rightarrow t^2={(x-1)}^{1/4}$
So, $y={(x-1)}^{5/4}+1$
On differentiating both sides with respect to $x$, we get
$\frac{dy}{dx}=\frac{5}{4}{(x-1)}^{1/4}$
Again, differentiating both sides with respect to $x$, we get
$\frac{d^{2}y}{d{x}^2}=\frac{5}{16}{(x-1)}^{-3/4}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{5}{16{(x-1)}^{3/4}}=\frac{5}{16{\left(t^2\right)}^3}=\frac{5}{16t^6}$