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

The tangent line to the curve at $P(1,\frac{\pi}{4})$ is $$y=\Big(\frac{\pi}{4}-\frac{1}{2}\Big)x+\frac{1}{2}$$
$$x+\tan\Big(\frac{y}{x}\Big)=2$$ The graph of the curve is enclosed below. a) As the point $(1,\frac{\pi}{4})$ lies in the curve, $P$ satisfies the equation. b) Find $dy/dx$ using implicit differentiation: $$1+\sec^2\Big(\frac{y}{x}\Big)\frac{d}{dx}\Big(\frac{y}{x}\Big)=0$$ $$1+\sec^2\Big(\frac{y}{x}\Big)\frac{x\frac{dy}{dx}-y}{x^2}=0$$ $$\frac{x^2+\sec^2(\frac{y}{x})(x\frac{dy}{dx}-y)}{x^2}=0$$ $$x^2+x\sec^2\Big(\frac{y}{x}\Big)\frac{dy}{dx}-y\sec^2\Big(\frac{y}{x}\Big)=0$$ $$x\sec^2\Big(\frac{y}{x}\Big)\frac{dy}{dx}=y\sec^2\Big(\frac{y}{x}\Big)-x^2$$ $$\frac{dy}{dx}=\frac{y\sec^2\Big(\frac{y}{x}\Big)-x^2}{x\sec^2(\frac{y}{x})}$$ - For $P(1,\frac{\pi}{4})$: $$\frac{dy}{dx}=\frac{\frac{\pi}{4}\sec^2(\frac{\pi}{4})-1^2}{1\times\sec^2(\frac{\pi}{4})}$$ We have $$\sec^2\frac{\pi}{4}=\frac{1}{\cos^2\frac{\pi}{4}}=\frac{1}{(\frac{\sqrt2}{2})^2}=\frac{1}{\frac{1}{2}}=2$$ Therefore, $$\frac{dy}{dx}=\frac{\frac{\pi}{4}\times2-1}{1\times2}=\frac{\frac{\pi}{2}-1}{2}=\frac{\pi}{4}-\frac{1}{2}$$ c) The tangent line to the curve at $P(1,\frac{\pi}{4})$ is $$y-\frac{\pi}{4}=\Big(\frac{\pi}{4}-\frac{1}{2}\Big)(x-1)$$ $$y-\frac{\pi}{4}=\Big(\frac{\pi}{4}-\frac{1}{2}\Big)x-\Big(\frac{\pi}{4}-\frac{1}{2}\Big)=\Big(\frac{\pi}{4}-\frac{1}{2}\Big)x-\frac{\pi}{4}+\frac{1}{2}$$ $$y=\Big(\frac{\pi}{4}-\frac{1}{2}\Big)x+\frac{1}{2}$$