If y-1a-z- then dzdy is-

The correct option is A (a z)^2 y=1a z d yd z= 1(a z)^2( 1) #160; #160;(By differentiating w.r.t z) d yd z=1(a z)^2 #160; #160; ...

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Conjugate of a Complex Number

If y=1a-z, ...

Question

If $y = \frac{1}{a - z}$ , then $\frac{d z}{d y}$ is:

(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

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\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = ?

\frac{x}{a} = \frac{y}{b} = \frac{z}{c},

\frac{x^{2} + a^{2}}{x + a} + \frac{y^{2} + b^{2}}{y + b} + \frac{z^{2} + c^{2}}{z + c} = \frac{(x + y + z)^{2} + (a + b + c)^{2}}{x + y + z + a + b + c} .

f (x, y, z) = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} - 1

Σ x \frac{\partial f}{\partial x}

x = a sec θ cos ϕ, y = b sec θ sin ϕ

z = c tan θ

(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = (1 + \frac{z^{2}}{c^{2}})

a, b, c

x, y, z

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1, - \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

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