# w = sqrt (4-2x^2-2y^2 )   , given x =r cosθ   and y = r sinθ, find dw/dr by the chain rule You should substitute `r cos theta`  for x and `r sin theta`  for y such that:

`w = sqrt(4 - 2x^2 - 2y^2) => w = sqrt(4 - 2(r^2 sin^2 theta + r^2 cos^2 theta))`

Factoring out `r^2 ` yields:

`w = sqrt(4 - 2r^2(sin^2 theta + cos^2 theta))`

Using...

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You should substitute `r cos theta`  for x and `r sin theta`  for y such that:

`w = sqrt(4 - 2x^2 - 2y^2) => w = sqrt(4 - 2(r^2 sin^2 theta + r^2 cos^2 theta))`

Factoring out `r^2 ` yields:

`w = sqrt(4 - 2r^2(sin^2 theta + cos^2 theta))`

Using fundamental formula of trigonometry yields:

`sin^2 theta + cos^2 theta = 1`

`w = sqrt(4 - 2r^2)`

You need to differentiate the function w with respect to r, using chain rule such that:

`(dw)/(dr) = (1/(2sqrt(4-2r^2)))*(d(4 - 2r^2))/(dr)`

`(dw)/(dr) = (1/(2sqrt(4-2r^2)))*(-4r)`

Reducing by 2 yields:

`(dw)/(dr) = ((-2r)/(sqrt(4-2r^2)))`

Hence, differentiating the given function with respect to r yields `(dw)/(dr) = ((-2r)/(sqrt(4-2r^2))).`

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