MHT CET · Maths · Differentiation
If \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4\), then \(\frac{d y}{d x}=\)
- A \(\frac{y-7 x}{7 x-y}\)
- B \(\frac{7 y-7 x}{y-7 x}\)
- C \(\frac{7 x+y}{x-7 y}\)
- D \(\frac{y+7 x}{7y-x}\)
Answer & Solution
Correct Answer
(B) \(\frac{7 y-7 x}{y-7 x}\)
Step-by-step Solution
Detailed explanation
(B)
Given : \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4\)
\(\therefore \frac{x+y}{\sqrt{x y}}=4 \Rightarrow x+y=4 \sqrt{x y}\)
Squaring both sides, we get,
\((x+y)^{2}=16 x y \Rightarrow x^{2}+y^{2}=14 x y\)
Differentiating both sides w.r.t. \(x\)
\(\begin{array}{l}
2 x+2 y \frac{d y}{d x}=14\left[x \frac{d y}{d x}+y\right] \Rightarrow x+y \frac{d y}{d x}=7 x \cdot \frac{d y}{d x}+7 y \\
(y-7 x) \frac{d y}{d x}=7 y-x \Rightarrow \frac{d y}{d x}=\frac{7 y-x}{y-7 x}
\end{array}\)
Given : \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4\)
\(\therefore \frac{x+y}{\sqrt{x y}}=4 \Rightarrow x+y=4 \sqrt{x y}\)
Squaring both sides, we get,
\((x+y)^{2}=16 x y \Rightarrow x^{2}+y^{2}=14 x y\)
Differentiating both sides w.r.t. \(x\)
\(\begin{array}{l}
2 x+2 y \frac{d y}{d x}=14\left[x \frac{d y}{d x}+y\right] \Rightarrow x+y \frac{d y}{d x}=7 x \cdot \frac{d y}{d x}+7 y \\
(y-7 x) \frac{d y}{d x}=7 y-x \Rightarrow \frac{d y}{d x}=\frac{7 y-x}{y-7 x}
\end{array}\)
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