TS EAMCET · Maths · Differentiation
If \(x=\frac{9 t^2}{1+t^4}\) and \(y=\frac{16 t^2}{1-t^4}\) then \(\frac{d y}{d x}=\)
- A \(\frac{16}{9}\left(\frac{1-t^4}{1+t^4}\right)^3\)
- B \(\frac{16\left(1-t^4\right)}{9\left(1+t^4\right)}\)
- C \(\frac{9\left(1-t^4\right)}{16\left(1+t^4\right)}\)
- D \(\frac{16}{9}\left(\frac{1+t^4}{1-t^4}\right)^3\)
Answer & Solution
Correct Answer
(D) \(\frac{16}{9}\left(\frac{1+t^4}{1-t^4}\right)^3\)
Step-by-step Solution
Detailed explanation
\(y=\frac{16 t^2}{1-t^4} ; \frac{d y}{d t}=\frac{\left(1-t^4\right) \times 32 t-16 t^2 \times\left(-4 t^3\right)}{\left(1-t^4\right)^2}\)…
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