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

$$\frac{dy}{dx}=\frac{xe^x+1}{x\sec^2y}$$
$$\tan y=e^x+\ln x$$ Use implicit differentiation here to find $dy/dx$. Differentiate both sides with respect to $x$: $$\frac{d}{dx}(\tan y)=e^x+\frac{1}{x}$$ Apply the Chain Rule for $(\tan y)$: $$\sec^2y\frac{dy}{dx}=e^x+\frac{1}{x}$$ $$\frac{dy}{dx}=\frac{e^x+\frac{1}{x}}{\sec^2y}=\frac{\frac{xe^x+1}{x}}{\sec^2y}$$ $$\frac{dy}{dx}=\frac{xe^x+1}{x\sec^2y}$$