if Xp Yq=(x+y)p+q,prove that dy_dx=y_x - Maths - Continuity and Differentiability

Question

if Xp Yq=(x+y)p+q,prove that dy_dx=y_x

Shridhar Kaushik · Accepted Answer

We have

$x^{p} y^{q} = (x + y)^{p + q}$

To Prove :

$\frac{d y}{d x} = \frac{y}{x}$

Since $x^{p} y^{q} = (x + y)^{p + q}$

Taking log on both sides, we get –

$\log x^{p} + \log y^{q} = (p + q) \log (x + y) \Rightarrow p \log x + q \log y = (p + q) \log (x + y)$

Differentiating with respect to x, we get –

$p \times \frac{1}{x} + q \times \frac{1}{y} \frac{d y}{d x} = (p + q) \frac{1}{(x + y)} (1 + \frac{d y}{d x}) \Rightarrow \frac{p}{x} + \frac{q}{y} \frac{d y}{d x} = \frac{(p + q)}{(x + y)} + (\frac{p + q}{x + y}) \frac{d y}{d x} \Rightarrow \frac{d y}{d x} [\frac{p + q}{x + y} - \frac{q}{y}] = \frac{p}{x} - (\frac{p + q}{x + y}) \Rightarrow \frac{d y}{d x} [\frac{p y + q y - q x - q y}{y (x + y)}] = \frac{p x + p y - p x - q x}{x (x + y)} \Rightarrow \frac{d y}{d x} \frac{(p y - q x)}{(y (x + y))} = \frac{(p y - q x)}{(x (x + y))} \Rightarrow \frac{d y}{d x} = \frac{y}{x} Hence Proved$

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