MHT CET · Maths · Differentiation
If \(\sqrt{x}+\sqrt{y}=\sqrt{x y}\), then \(\frac{d y}{d x}=\)
- A \(-\left(\frac{y}{x}\right)^{\frac{3}{2}}\)
- B \(\left(\frac{x}{y}\right)^{\frac{3}{2}}\)
- C \(-\left(\frac{x}{y}\right)^{\frac{3}{2}}\)
- D \(\left(\frac{y}{x}\right)^{\frac{3}{2}}\)
Answer & Solution
Correct Answer
(A) \(-\left(\frac{y}{x}\right)^{\frac{3}{2}}\)
Step-by-step Solution
Detailed explanation
We have \(\sqrt{x}+\sqrt{y}=\sqrt{x y}\)
Dividing both sides by \(\sqrt{x y}\), we get
\(\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}=1\)
Differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
&\left(\frac{-1}{2}\right)\left(\frac{1}{y^{3 / 2}}\right)\left(\frac{d y}{d x}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{x^{3 / 2}}\right)=0 \\
\therefore & \frac{d y}{d x}=-\left(\frac{y}{x}\right)^{3 / 2}
\end{aligned}\)
Dividing both sides by \(\sqrt{x y}\), we get
\(\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}=1\)
Differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
&\left(\frac{-1}{2}\right)\left(\frac{1}{y^{3 / 2}}\right)\left(\frac{d y}{d x}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{x^{3 / 2}}\right)=0 \\
\therefore & \frac{d y}{d x}=-\left(\frac{y}{x}\right)^{3 / 2}
\end{aligned}\)
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