MHT CET · Maths · Differentiation
If \(x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}, x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\)
- A \(\frac{1}{x^3 y}\)
- B \(\frac{1}{x y^3}\)
- C \(-\frac{1}{x y^3}\)
- D \(-\frac{1}{x^3 y}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{x^3 y}\)
Step-by-step Solution
Detailed explanation
\(x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}} \text { and } x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2} \)
\( \therefore \left(x^2+y^2\right)^2=\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^2 \)
\( \therefore x^4+y^4+2 x^2 y^2=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 \)
\( \therefore \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 \)
\( \therefore x^2 \dot{y}^2=1 \)
\( \therefore y^2=\frac{1}{x^2}\)
Differentiating w.r.t. \(x\), we get
\(2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-2}{x^3} \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-1}{x^3 y}\)
\( \therefore \left(x^2+y^2\right)^2=\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^2 \)
\( \therefore x^4+y^4+2 x^2 y^2=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 \)
\( \therefore \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 \)
\( \therefore x^2 \dot{y}^2=1 \)
\( \therefore y^2=\frac{1}{x^2}\)
Differentiating w.r.t. \(x\), we get
\(2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-2}{x^3} \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-1}{x^3 y}\)
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