Guys, is this right?

And if it is, from the the y' got in there?

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- Aug 18th 2008, 09:24 AMasi123Derivative
Guys, is this right?

And if it is, from the the y' got in there? - Aug 18th 2008, 09:29 AMo_O
$\displaystyle \begin{array}{lcl}z & = & \sin \frac{1}{y} \\ z' & = & \underbrace{\left(\cos \frac{1}{y}\right) \cdot \left(\frac{1}{y}\right)'}_{\text{By chain rule}} \end{array}$

etc. etc.

y' shouldn't be there. - Aug 18th 2008, 10:00 AMwingless
y' should be there only if you are differentiating with respect to a variable except y.