What is the derivative of #w =sqrt(x^2+y^2+z^2)#?

1 Answer
Aug 30, 2015

Answer:

#(delw)/(delx) = x/sqrt(x^2 + y^2 + z^2)#
#(delw)/(dely) = y/sqrt(x^2 + y^2 + z^2)#
#(delw)/(delz) = z/sqrt(x^2 + y^2 + z^2)#

Explanation:

Since you're dealing with a multivariable function, you must treat #x#, #y#, and #z# as independent variables and calculate the partial derivative of #w#, your dependent variable, with respect to #x#, #y#, and #z#.

When you differentiate with respect to #x#, you treat #y# and #z# as constants. LIkewise, when you differentiate with respect to #y#, you treat #x# and #z# as constants, and so on.

So, let's first differentiate #w# with respect to #x#

#(delw)/(delx) = (del)/(delx)sqrt(x^2 + y^2 + z^2)#

#(delw)/(delx) = 1/2 * (x^2 + y^2 + z^2) ^(-1/2) * (del)/(delx)(x^2 + y^2 + z^2)#

#(delw)/(delx) = 1/color(red)(cancel(color(black)(2))) * 1/sqrt(x^2 + y^2 + z^2) * (color(red)(cancel(color(black)(2)))x + 0 + 0)#

#(delw)/(delx) = color(green)(x/sqrt(x^2 + y^2 + z^2))#

You don't need to calculate the other two partial derivatives, all you have to do is recognize that the only thing that changes when you differentiate with respect to #y# is that you get

#(del)/(dely)(x^2 + y^2 + z^2) = 0 + 2y + 0#

The same is true for the deivative with respect to #z#

#(del)/(delz)(x^2 + y^2 + z^2) = 0 + 0 + 2z#

This means that you have

#(delw)/(dely) = color(green)(y/sqrt(x^2 + y^2 + z^2))#

and

#(delw)/(delx) = color(green)(z/sqrt(x^2 + y^2 + z^2))#