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January 29, 2015

January 29, 2015

Posted by **Anonymous** on Wednesday, October 14, 2009 at 3:42pm.

= 2yw/240^2 + 0 + y^2/240^2

but final answer is....

= 2yw/240^2

So, is the question asking for partial derivative or first derivative? I think partial because the final answer only took the derivative with respect to y.

- Is this f'(x) or partial derivative? -
**MathMate**, Wednesday, October 14, 2009 at 5:19pmIf VC(y)=((y/240)^2)(w)

then VC is not dependent on w, so if you differentiate as a product, the term containing dw/dy drops out (because dw/dy=0), giving you the correct answer.

If w is a function of y, namely w=w(y), then the term dw/dy should be kept:

d(VC)/dy = 2yw/240^2 + y^2/240^2*(dw/dy)

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