MHT CET · Maths · Differentiation
If \(y\) is a function of \(x\) and \(\log (x+y)=2 x y\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=0\) is
- A \(0\)
- B \(-1\)
- C \(1\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
\(\log (x+y)=2 x y...(i)\)
Differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
& \left(\frac{1}{x+y}\right)\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2\left(x \frac{\mathrm{d} y}{\mathrm{~d} x}+y\right) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1-2 x y-2 y^2}{2 x^2+2 x y-1}
\end{aligned}\)
Putting \(x=0\) in (i), we get
\(\begin{aligned}
& y=1 \\
\therefore \quad & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\frac{1-0-2}{0+0-1}=1
\end{aligned}\)
Differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
& \left(\frac{1}{x+y}\right)\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2\left(x \frac{\mathrm{d} y}{\mathrm{~d} x}+y\right) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1-2 x y-2 y^2}{2 x^2+2 x y-1}
\end{aligned}\)
Putting \(x=0\) in (i), we get
\(\begin{aligned}
& y=1 \\
\therefore \quad & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\frac{1-0-2}{0+0-1}=1
\end{aligned}\)
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