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 \(y\)!
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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!
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Finally, all we need to do is solve this for \(y'\).
\[\require{bbox} \bbox[2pt,border:1px solid black]{{y' = \frac{{1 - {{\bf{e}}^x}}}{{ - \cos \left( y \right)}} = \left( {{{\bf{e}}^x} - 1} \right)\sec \left( y \right)}}\]