MCQ
Let $y=t^{10}+1$ and $x=t^8+1$, then $\frac{d^2 y}{d x^2}$ is equal to
- A$\frac{5}{2} t$
- ✓$\frac{5}{16 t^6}$
- C$20 t^8$
- DNone of these
✓
Answer
Correct option: B.
$\frac{5}{16 t^6}$
We have, $y=t^{10}+1, x=t^8+1$
$\Rightarrow \frac{d y}{d t}=10 t^9, \frac{d x}{d t}=8 t^7$
$\therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{10 t^9}{8 t^7}=\frac{5}{4} t^2$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{5}{4}(2 t) \frac{d t}{d x}$
$=\frac{5}{4} \times 2 t \times \frac{1}{8 t^7}$
$=\frac{5}{16 t^6}$
$\Rightarrow \frac{d y}{d t}=10 t^9, \frac{d x}{d t}=8 t^7$
$\therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{10 t^9}{8 t^7}=\frac{5}{4} t^2$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{5}{4}(2 t) \frac{d t}{d x}$
$=\frac{5}{4} \times 2 t \times \frac{1}{8 t^7}$
$=\frac{5}{16 t^6}$
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