## University Calculus: Early Transcendentals (3rd Edition)

$$\frac{dy}{dx}=\cos y\cot y$$
$$x=\sec y$$ 1) Differentiate both sides of the equation with respect to $x$: $$\frac{dx}{dx}=\frac{d}{dx}(\sec y)$$ $$1=\sec y\tan y\frac{dy}{dx}$$ 2) Now solve for $dy/dx$: $$\frac{dy}{dx}=\frac{1}{\sec y\tan y}=\frac{1}{\sec y}\times\frac{1}{\tan y}$$ $$\frac{dy}{dx}=\frac{1}{\frac{1}{\cos y}}\times\cot y=\cos y\cot y$$