Question

The partial differential equation of the family of surfaces z=x+y+f(xy) ised​

Answer 1

Explanation:

Consider g = xy

$g_{x} = y\\g_{y} = x$

First partials are

∂

$z_{x} = \frac{∂}{∂x} (x + y + f(g))$

    = 1 + $f^{'}(g)g_{x}$

    = $1 + xf^{'} (g)$

$z_{y} = 1 + xf^{'}\\xf^{'} = z_{y} - 1\\f^{'} = \frac{z_{y} - 1}{x}$

$z_{x} = 1 + yf^{'}\\ z_{x} = 1 + y (\frac{z_{y}-1 }{x} )\\xz_{x} = x + yz_{y} - y\\xz_{x} - yz_{y} = x- y$

Answer 2

Answer:

Step-by-step explanation:

Concept :  Partial derivative of a function means that its derivative with respect to one variable, with the other held constant.

To find : Partial derivative of z = x + y + f(xy)

Solution :

Consider, p = xy

$p_{x} = y$

$p_{y} = x$

$z_{x} = \frac{d}{dx} (x + y + f(p))$

    $= 1 + f'(p)p_{x}$

    $= 1 + xf'(p)$

$z_{y} = \frac{d}{dy}(x + y + f(p))$

    $= 1 + f'(p)p_{x}$

    $= 1 + yf'(p)$

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