-y-x-x-y5- then d y-d x-

The correct option is C y/x nbsp; nbsp;√(y)√(x)= √((x+y)^5) ⇒ y^2x^3=(x+y)^5 Taking log both sides, we get 2 log y + 3 log x =5 log (x+y) Differentiating w ...

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Standard XII

Mathematics

Logarithmic Differentiation

√y√x=√x+y5, t...

Question

$\sqrt[3]{y} \sqrt{x} = \sqrt[6]{(x + y)^{5}}$ , then $\frac{d y}{d x} =$

x - y

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\frac{x}{y}

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\frac{y}{x}

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x + y

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Solution

The correct option is C $\frac{y}{x}$
$\sqrt[3]{y} \sqrt{x} = \sqrt[6]{(x + y)^{5}}$
$\Rightarrow y^{2} x^{3} = (x + y)^{5}$
Taking log both sides, we get
$2 log y + 3 log x = 5 log (x + y)$
Differentiating wrt $x$ ,
$\frac{2}{y} \frac{d y}{d x} + \frac{3}{x} = \frac{5}{x + y} (1 + \frac{d y}{d x})$
$\Rightarrow \frac{d y}{d x} = \frac{y}{x}$

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Catwoman says.
1) If

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3√y√x=6√(x+y)5, then dydx=

The correct option is C yx 3√y√x=6√(x+y)5 ⇒y2x3=(x+y)5 Taking log both sides, we get 2logy+3logx=5log(x+y) Differentiating wrt x, 2ydydx+3x=5x+y(1+dydx) ⇒dydx=yx

$\sqrt[3]{y} \sqrt{x} = \sqrt[6]{(x + y)^{5}}$ , then $\frac{d y}{d x} =$