Question
Find d VC/ d y = ((y/240)^2)(w)
= 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.
= 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.
Answers
If 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)
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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