COMEDK · Maths · 26. Differentiation
If \(x=\frac{1-t}{1+t}, y=\frac{2 t}{1+t}\), then \(\frac{d^{2} y}{d x^{2}}=\)
- A \(\frac{2 t}{(1+t)^{2}}\)
- B \(\frac{1}{(1+t)^{4}}\)
- C \(\frac{2 t}{(1+t)^{2}}\)
- D 0
Answer & Solution
Correct Answer
(D) 0
Step-by-step Solution
Detailed explanation
We have, \(x=\frac{1-t}{1+t}\) \[ \frac{d x}{d t}=\frac{-(1+t)-(1-t)}{(1+t)^{2}}=\frac{-2}{1+t^{2}} \text { and } y=\frac{2 t}{1+t} \] \[ \frac{d y}{d t}=\frac{(1+t) 2-2 t}{(1+t)^{2}}=\frac{2}{(1+t)^{2}} \] Now, \(\quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=-1\) and…
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