Guys, is this right? And if it is, from the the y' got in there?
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$\displaystyle \begin{array}{lcl}z & = & \sin \frac{1}{y} \\ z' & = & \underbrace{\left(\cos \frac{1}{y}\right) \cdot \left(\frac{1}{y}\right)'}_{\text{By chain rule}} \end{array}$ etc. etc. y' shouldn't be there.
y' should be there only if you are differentiating with respect to a variable except y.
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