Using Implicit Differentiation

1. Find \(y^\prime \) if \(e^{y}\cos x=x+1\).
2. Find an equation of the tangent line to the graph at the point \((0,0) \).

Solution (a) We use implicit differentiation: \[ \begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{d}{dx}( e^{y}\cos x) &= \dfrac{d}{dx}(x+1) \\ e^{y}\cdot \dfrac{d}{dx}( \cos x) +\left( \dfrac{d}{dx} e^{y}\right) \cdot \cos x &= 1 & {\color{#0066A7}{\hbox{Use the Product Rule.}}} \\ e^{y}( -\sin x) +e^{y}y^\prime \cdot \cos x &= 1 \\ ( e^{y}\cos x) y^\prime &= 1+e^{y}\sin x \\ y^\prime &= \dfrac{1+e^{y}\sin x}{e^{y}\cos x} \end{array} \end{eqnarray*} \]

(b) At the point \((0,0) \), the derivative \(y^\prime \) is \(\dfrac{1+e^{0}\sin 0}{e^{0}\cos 0}=\dfrac{1+1\cdot 0}{1\cdot 1}=1\), so the slope of the tangent line to the graph at \((0,0) \) is \(1\). An equation of the tangent line to the graph at the point \((0,0)\) is \(y=x.\)