MHT CET · Maths · Differentiation
If \(x^{2}+y^{2}=t+\frac{1}{t}, x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}\) then \(\frac{d y}{d x}=\)
- A \(-\frac{y}{x}\)
- B \(\frac{y}{x}\)
- C \(\frac{x}{2 y}\)
- D \(-\frac{x}{2 y}\)
Answer & Solution
Correct Answer
(A) \(-\frac{y}{x}\)
Step-by-step Solution
Detailed explanation
(A)
We have,
\(x^{2}+y^{2}=t+\frac{1}{t}$$\cdots\) (1) \(\quad\) and \(\quad x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}\)...(2)
Squaring (1), we get
\(x^{4}+y^{4}+2 x^{2} y^{2}=t^{2}+\frac{1}{t^{2}}+2 \Rightarrow x^{4}+y^{4}+2 x^{2} y^{2}=\) \(x^{4}+y^{4}+2 \text {..(from (2)) } \)
\( \therefore x^{2} y^{2}=1\)
Differentiating w.r.t. \(x\), we get
\(x^{2}\left(2 y \frac{d y}{d x}\right)+y^{2}(2 x)=0\)
\(\therefore \frac{d y}{d x}=\frac{-2 x y^{2}}{2 x^{2} y}=\frac{-y}{x}\)
We have,
\(x^{2}+y^{2}=t+\frac{1}{t}$$\cdots\) (1) \(\quad\) and \(\quad x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}\)...(2)
Squaring (1), we get
\(x^{4}+y^{4}+2 x^{2} y^{2}=t^{2}+\frac{1}{t^{2}}+2 \Rightarrow x^{4}+y^{4}+2 x^{2} y^{2}=\) \(x^{4}+y^{4}+2 \text {..(from (2)) } \)
\( \therefore x^{2} y^{2}=1\)
Differentiating w.r.t. \(x\), we get
\(x^{2}\left(2 y \frac{d y}{d x}\right)+y^{2}(2 x)=0\)
\(\therefore \frac{d y}{d x}=\frac{-2 x y^{2}}{2 x^{2} y}=\frac{-y}{x}\)
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