How do you find the derivative of sqrt(x^2+y^2)x2+y2?

1 Answer
Jun 23, 2015

To find d/dx(sqrt(x^2+y^2))ddx(x2+y2), as part of an implicit differentiation problem, use the chain rule.

Explanation:

d/dx(sqrtx) = 1/(2sqrtx)ddx(x)=12x, so

d/dx(sqrtu) = 1/(2sqrtu) (du)/dxddx(u)=12ududx.

d/dx(sqrt(x^2+y^2)) = 1/(2sqrt(x^2+y^2)) * d/dx(x^2+y^2)ddx(x2+y2)=12x2+y2ddx(x2+y2)

= 1/(2sqrt(x^2+y^2))(2x+2y dy/dx)=12x2+y2(2x+2ydydx)

=1/(2sqrt(x^2+y^2))2x+ 1/(2sqrt(x^2+y^2))2y dy/dx =12x2+y22x+12x2+y22ydydx

=x/sqrt(x^2+y^2)+ y/sqrt(x^2+y^2) dy/dx =xx2+y2+yx2+y2dydx

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