Partial differentiation is fun!

Calculus Level 4

If $z(x+y) = x^2 + y^2$ , then what is the value of

$\left(\dfrac{\partial z}{\partial x} - \dfrac{\partial z}{\partial y} \right)^2 ?$ For more problems on mathematics, click here.

4\left(1-\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)

4\left(1+\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)

2\left(1-\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)

4\left(1-\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\right)

4\left(-1-\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)

1 solution

Tom Engelsman
Oct 29, 2016

If z = (x^2 + y^2)/(x+y), then:

dz/dx = [(2x)*(x+y) - (x^2 + y^2)]/(x+y)^2 = (x^2 + 2xy - y^2)/(x+y)^2 (i);

dz/dy = [(2y)*(x+y) - (x^2 + y^2)]/(x+y)^2 = (y^2 + 2xy - x^2)/(x+y)^2 (ii);

dz/dx - dz/dy = (2x^2 - 2y^2)/(x+y)^2 = 2 (x-y)/(x+y) = 2 [1 - 2y/(x+y)] (iii).

Now squaring (iii) produces:

(dz/dx - dz/dy)^2 = 4*[1 - 4y/(x+y) + (4y^2)/(x+y)^2];

or 4*[1 - 4y(x+y)/(x+y)^2 + (4y^2)/(x+y)^2];

or 4*[1 + (4y^2 - 4xy - 4y^2)/(x+y)^2];

or 4*[1 - 4xy/(x+y)^2];

or 4*[1 - {(x^2 + 2xy - y^2)/(x+y)^2} - {(y^2 + 2xy - x^2)/(x+y)^2}];

or 4*(1 - dz/dx - dz/dy)

thus, choice B is correct.

Good solution! You should learn LaTeX for better solutions.

Sahil Silare - 4 years, 7 months ago