Question

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

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

Answer 1

Answer: (C)

Here $y=t^{10}+1$ and $x=t^8+1$
$t^8=x-1$
$\Rightarrow t^2=(x-1{)}^{1/4}$
So, $y=(x-1{)}^{5/4}+1$
Differentiate both sides w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{5}{4}(x-1{)}^{1/4}$
Again, differentiate both sides w.r.t. x, we get
$\frac{d^{2}y}{d{x}^2}=\frac{5}{16}(x-1{)}^{-3/4}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{5}{16(x-1{)}^{3/4}}$
$=\frac{5}{16{\left(t^2\right)}^3}=\frac{5}{16t^6}$