If x^{p}y^{q}=(x+y)^{p+q} then Prove that `dy/dx=y/x`

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#### Solution

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

Taking log both side

`p log x+ q log y=( p+q)log(x +y)`

Diff. w.r.t. x

`p/x+q/ydy/dx=(p+q)/(x+y)+((p+q)/(x+y))dy/dx`

`q/ydy/dx-((p+q)/(x+y)) dy/dx=(p+q)/(x+y)-p/x`

`(q/y-(p+q)/(x+y)) dy/dx=((p+q)/(x+y)-p/x)`

`((qx-py)/y)dy/dx=((qx-py)/x)`

`1/y dy/dx=1/x`

`dy/dx=y/x`

Concept: Exponential and Logarithmic Functions

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