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Section 3.10 : Implicit Differentiation
6. Find \(y'\) by implicit differentiation for \({{\bf{e}}^x} - \sin \left( y \right) = x\).
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Hint : Don’t forget that \(y\) is really \(y\left( x \right)\) and so we’ll need to use the Chain Rule when taking the derivative of terms involving \(y\)!
First, we just need to take the derivative of everything with respect to \(x\) and we’ll need to recall that \(y\) is really \(y\left( x \right)\) and so we’ll need to use the Chain Rule when taking the derivative of terms involving \(y\).
Differentiating with respect to \(x\) gives,
\[{{\bf{e}}^x} - \cos \left( y \right)y' = 1\]Don’t forget the \(y'\) on the cosine after differentiating. Again, \(y\) is really \(y\left( x \right)\) and so when differentiating \(\sin \left( y \right)\) we really differentiating \(\sin \left[ {y\left( x \right)} \right]\) and so we are differentiating using the Chain Rule!
Show Step 2Finally, all we need to do is solve this for \(y'\).
\[\require{bbox} \bbox[2pt,border:1px solid black]{{y' = \frac{{1 - {{\bf{e}}^x}}}{{ - \cos \left( y \right)}} = \left( {{{\bf{e}}^x} - 1} \right)\sec \left( y \right)}}\]