The correct option is A (a z)^2 y = ( 1 a z #160; ) Let a z = t ⇒ dt dz = 1 ∴ y = 1t ⇒dydt = 1t^2 ∴dydz = dydt×dtdz ...

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Parametric Differentiation

If y=1a-z, ...

Question

If $y = \frac{1}{a - z}$ , find $\frac{d z}{d y}$

(a - z)^{2}

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- (z - a)^{2}

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(z + a)^{2}

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- (z + a)^{2}

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Solution

The correct option is A $(a - z)^{2}$
$y = (\frac{1}{a - z})$
Let $a - z = t$

$\Rightarrow \frac{d t}{d z} = - 1$
$∴ y = \frac{1}{t}$

$\Rightarrow \frac{d y}{d t} = \frac{- 1}{t^{2}}$

$∴ \frac{d y}{d z} = \frac{d y}{d t} \times \frac{d t}{d z}$

$= \frac{- 1}{t^{2}} . - 1$

$= \frac{1}{t^{2}}$

$\Rightarrow \frac{d z}{d y} = t^{2}$

$= {(a - z)}^{2}$

$∴ \frac{d z}{d y} = {(a - z)}^{2} .$

Hence the answer is ${(a - z)}^{2} .$

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