MHT CET · Maths · Differentiation
If \(x^{p} y^{q}=(x+y)^{p+q}\), then \(\frac{d y}{d x}\) is equal to
- A \(y / x\)
- B \(p y / q x\)
- C \(x / y\)
- D \(q y / p x\)
Answer & Solution
Correct Answer
(A) \(y / x\)
Step-by-step Solution
Detailed explanation
Given, \(x^{p} y^{q}=(x+y)^{p+q}\)
Taking log on both sides, we get
\(
\begin{array}{l}
p \log x+q \log y=(p+q) \log (x+y) \\
\Rightarrow\frac{p}{x}+\frac{q}{y} \frac{d y}{d x}=\frac{(p+q)}{(x+y)}\left(1+\frac{d y}{d x}\right) \\
\Rightarrow \left(\frac{p}{x}-\frac{p+q}{x+y}\right)=\left(\frac{p+q}{x+y}-\frac{q}{y}\right) \frac{d y}{d x} \\
\Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{array}
\)
Taking log on both sides, we get
\(
\begin{array}{l}
p \log x+q \log y=(p+q) \log (x+y) \\
\Rightarrow\frac{p}{x}+\frac{q}{y} \frac{d y}{d x}=\frac{(p+q)}{(x+y)}\left(1+\frac{d y}{d x}\right) \\
\Rightarrow \left(\frac{p}{x}-\frac{p+q}{x+y}\right)=\left(\frac{p+q}{x+y}-\frac{q}{y}\right) \frac{d y}{d x} \\
\Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{array}
\)
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