MHT CET · Maths · Differentiation
If \(\log (x+y)=\sin (x+y)\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) is
- A 2
- B 1
- C 0
- D -1
Answer & Solution
Correct Answer
(D) -1
Step-by-step Solution
Detailed explanation
\(\log (x+y)=\sin (x+y)\)
Differentiating both sides w.r.t: \(x\), we get
\(\begin{aligned}
& \frac{1}{x+y}\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=\cos (x+y)\left[1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right] \\
& \Rightarrow \frac{1}{x+y}+\frac{1}{x+y} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=\cos (x+y)+\cos (x+y) \frac{\mathrm{d} y}{\mathrm{~d} x} \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}\left(\frac{1}{x+y}-\cos (x+y)\right)=\cos (x+y)-\frac{1}{x+y} \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=-1
\end{aligned}\)
Differentiating both sides w.r.t: \(x\), we get
\(\begin{aligned}
& \frac{1}{x+y}\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=\cos (x+y)\left[1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right] \\
& \Rightarrow \frac{1}{x+y}+\frac{1}{x+y} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=\cos (x+y)+\cos (x+y) \frac{\mathrm{d} y}{\mathrm{~d} x} \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}\left(\frac{1}{x+y}-\cos (x+y)\right)=\cos (x+y)-\frac{1}{x+y} \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=-1
\end{aligned}\)
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