(x+y)^{a+b} = x^ay^b ⇒ \frac{dy}{dx} =

TS ICET 23 May 2019 Shift 2 Paper

Question : 135 of 200

Marks: +1, -0

(x+y)^{a+b}

x^ay^b

⇒

{dy}/{dx}

Solution:

(x+y)^{a+b}=x^{a} y^{b}

(a+b) \log (x+y)=a \log x+b \log y

Differentiating both sides w.r.to x

{a+b}/{x+y}(1+\;{d y}/{d x})={a}/{x}+{b}/{y} ⋅{d y}/{d x}

\;{d y}/{d x}(\;{a+b}/{x+y}-\;{b}/{y})=(\;{a}/{x}-\;{a+b}/{x+y})

\;{d y}/{d x}(\;{a y+b y-b x-b y}/{y(x+y)})=(\;{a x+a y-a x-b x}/{x(x+y)})

\;{d y}/{d x}=\;{y}/{x}

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