If xan-ybn-Zcn-1-then - z- x-here z is a function of x and y

The correct option is A (c/a)^n(xz)^n 1 Given #160; (xa)^n+(yb)^n+(zc)^n=1 #160; #160; #160; #160; #160; #160; here z is a func ...

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Standard XII

Mathematics

Range

If xan+ybn+...

Question

$I$ f ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} + {(\frac{Z}{c})}^{n} = 1$ , then $\frac{\partial z}{\partial x} =$
(here $z$ is a function of $x$ and $y$ )

- {(\frac{c}{a})}^{n} {(\frac{x}{z})}^{n - 1}

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- {(\frac{a}{c})}^{n} {(\frac{z}{x})}^{n - 1}

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- {(\frac{a}{b})}^{n} {(\frac{x}{y})}^{n - 1}

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- {(\frac{b}{c})}^{n} {(\frac{y}{x})}^{n - 1}

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Solution

The correct option is A $- {(\frac{c}{a})}^{n} {(\frac{x}{z})}^{n - 1}$
Given ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} + {(\frac{z}{c})}^{n} = 1$ here $z$ is a function of $x$ and $y$

applying partial derivative on both sides w.r.t $x$ gives

$\frac{n}{a} {(\frac{x}{a})}^{n - 1} + 0 + \frac{n}{c} {(\frac{z}{c})}^{n - 1} \frac{\partial z}{\partial x} = 0$

$\Rightarrow \frac{\partial z}{\partial x} = \frac{- \frac{n}{a} {(\frac{x}{a})}^{n - 1}}{\frac{n}{c} {(\frac{z}{c})}^{n - 1}} = - {(\frac{c}{a})}^{n} {(\frac{x}{z})}^{n - 1}$

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If (xa)n+(yb)n+(Zc)n=1, then ∂z∂x= (here z is a function of x and y)

The correct option is A −(ca)n(xz)n−1Given (xa)n+(yb)n+(zc)n=1 here z is a function of x and yapplying partial derivative on both sides w.r.t x givesna(xa)n−1+0+nc(zc)n−1∂z∂x=0⇒∂z∂x=−na(xa)n−1nc(zc)n−1=−(ca)n(xz)n−1

$I$ f ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} + {(\frac{Z}{c})}^{n} = 1$ , then $\frac{\partial z}{\partial x} =$
(here $z$ is a function of $x$ and $y$ )