# Given z = ln(sqrt x^2 + y^2), show that x dz/dx  + y dz/dy = 1 Calculus of several variables question

```(dz)/(dx) = 1/sqrt(x^2 + y^2) (d/(dx) sqrt(x^2 + y^2))`

since `d/(dx) ln(f(x)) = (f'(x))/f(x)`

`d/(dx) sqrt(x^2 + y^2) = (1/2)(x^2+y^2)^(-1/2)(2x) = x/sqrt(x^2+y^2)`

So we have

`(dz)/(dx) = 1/(sqrt(x^2+y^2))(x/sqrt(x^2+y^2)) = x/(x^2+y^2)`

By the symmetry of the function `z = ln(sqrt(x^2+y^2))` ,

`(dz)/(dy) = y/(x^2+y^2)`

`therefore`   `y ((dz)/(dx)) + y((dz)/(dy)) = x^2/(x^2+y^2)...

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```(dz)/(dx) = 1/sqrt(x^2 + y^2) (d/(dx) sqrt(x^2 + y^2))`

since `d/(dx) ln(f(x)) = (f'(x))/f(x)`

`d/(dx) sqrt(x^2 + y^2) = (1/2)(x^2+y^2)^(-1/2)(2x) = x/sqrt(x^2+y^2)`

So we have

`(dz)/(dx) = 1/(sqrt(x^2+y^2))(x/sqrt(x^2+y^2)) = x/(x^2+y^2)`

By the symmetry of the function `z = ln(sqrt(x^2+y^2))` ,

`(dz)/(dy) = y/(x^2+y^2)`

`therefore`   `y ((dz)/(dx)) + y((dz)/(dy)) = x^2/(x^2+y^2) + y^2/(x^2+y^2) = 1`

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