TS EAMCET · Maths · Differentiation
If \(x=\frac{\mathrm{t}^2}{1+\mathrm{t}^5}, y=\frac{2 \mathrm{t}^3}{1+\mathrm{t}^5}\) and \(\mathrm{t} \neq-1\) is a parameter then \(\frac{d y}{d x}=\)
- A \(\frac{2\left(3+2 t^5\right)}{\left(2-3 t^5\right)}\)
- B \(\frac{2 t\left(3-2 t^5\right)}{\left(2-3 t^5\right)}\)
- C \(\frac{2 t\left(3-2 t^5\right)}{\left(2+3 t^5\right)}\)
- D \(\frac{2\left(3+2 t^5\right)}{\left(2+3 t^5\right)}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 t\left(3-2 t^5\right)}{\left(2-3 t^5\right)}\)
Step-by-step Solution
Detailed explanation
\(\frac{dx}{dt} = \frac{2t(1+t^5) - t^2(5t^4)}{(1+t^5)^2} = \frac{2t+2t^6-5t^6}{(1+t^5)^2} = \frac{t(2-3t^5)}{(1+t^5)^2}\) \(\frac{dy}{dt} = \frac{6t^2(1+t^5) - 2t^3(5t^4)}{(1+t^5)^2} = \frac{6t^2+6t^7-10t^7}{(1+t^5)^2} = \frac{2t^2(3-2t^5)}{(1+t^5)^2}\)…
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