Question

How do you find the derivative of `sqrt(x^2+y^2)`?

Answer 1

To find `d/dx(sqrt(x^2+y^2))`, as part of an implicit differentiation problem, use the chain rule.

Explanation:

`d/dx(sqrtx) = 1/(2sqrtx)`, so

`d/dx(sqrtu) = 1/(2sqrtu) (du)/dx`.

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

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

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

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

In order to solve for `dy/dx` you will, of course, need the rest of the derivative of the rest of the original equation.