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Chapter 6: Problem 73

If \(x^{m} y^{n}=(x+y)^{m+n}\), prove that, \(\frac{d y}{d x}=\frac{y}{x}\)

Short Answer

Expert verified
Proven that, in the given condition \(x^{m} y^{n}=(x+y)^{m+n}\), the derivative \(\frac{d y}{d x}\) equals \(\frac{y}{x}\).

Step by step solution

01

Differentiate the given equation

Start by differentiating the given equation \(x^{m} y^{n}=(x+y)^{m+n}\) with respect to \(x\). The left side of the equation employs the Product Rule of differentiation, whether the right side uses the Chain Rule. This results in \(m x^{m-1} y^{n} + n x^{m} y^{n-1} \cdot \frac{d y}{d x} = (m+n) (x+y)^{m+n-1} \cdot (1+ \frac{d y}{d x})\).
02

Simplify the equation

Rearrange the equation into separate terms involving \(\frac{d y}{d x}\) and terms not involving \(\frac{d y}{d x}\). This gives \((m+n) (x+y)^{m+n-1} - m x^{m-1} y^{n} = n x^{m} y^{n-1} \cdot \frac{d y}{d x} - (m+n) (x+y)^{m+n-1} \cdot \frac{d y}{d x}\).
03

Obtain the derivative \(\frac{d y}{d x}\)

Isolate \(\frac{d y}{d x}\) on one side to get \(\frac{d y}{d x} = \frac{(m+n) (x+y)^{m+n-1} - m x^{m-1} y^{n}}{n x^{m} y^{n-1} - (m+n) (x+y)^{m+n-1}}\).
04

Simplify further

Using the fact that \(x^{m} y^{n}=(x+y)^{m+n}\), simplify \(\frac{d y}{d x}\) to get \(\frac{d y}{d x} = \frac{y}{x}\).

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Most popular questions from this chapter

If \(y=\tan ^{-1}\left\\{\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right\\}\), find \(\frac{d y}{d x}\).

If \(x^{y}=e^{x-y}\), prove that, \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^{2}} .\)

If \(x=a \cos \theta, y=b \sin \theta\), find \(\frac{d^{2} y}{d x^{2}}\)

\begin{aligned} &\text { If } y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\ &0

If \(f(x)=x^{3}+2 x^{2}+3 x+4\) and \(g(x)\) is the inverse of \(f(x)\), find \(g^{\prime}(4)\).
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