Answer
The tangent line to the curve at $P(\frac{\pi}{4},0)$ is $$y=-x+\frac{\pi}{4}$$
Work Step by Step
$$xy^3+\tan(x+y)=1$$
The graph of the curve is enclosed below.
a) As the point $(\frac{\pi}{4},0)$ lies in the curve, $P$ satisfies the equation.
b) Find $dy/dx$ using implicit differentiation: $$(y^3+3xy^2\frac{dy}{dx})+\sec^2(x+y)\frac{d}{dx}(x+y)=0$$ $$y^3+3xy^2\frac{dy}{dx}+\sec^2(x+y)(1+\frac{dy}{dx})=0$$ $$y^3+3xy^2\frac{dy}{dx}+\sec^2(x+y)+\sec^2(x+y)\frac{dy}{dx}=0$$ $$\frac{dy}{dx}(3xy^2+\sec^2(x+y))=-(y^3+\sec^2(x+y))$$ $$\frac{dy}{dx}=-\frac{y^3+\sec^2(x+y)}{3xy^2+\sec^2(x+y)}$$
- For $P(\frac{\pi}{4},0)$: $$\frac{dy}{dx}=-\frac{0^3+\sec^2(\frac{\pi}{4}+0)}{3\times\frac{\pi}{4}\times0^2+\sec^2(\frac{\pi}{4}+0)}=-\frac{\sec^2\frac{\pi}{4}}{0+\sec^2\frac{\pi}{4}}=-1$$
c) The tangent line to the curve at $P(\frac{\pi}{4},0)$ is $$y-0=-(x-\frac{\pi}{4})$$ $$y=-x+\frac{\pi}{4}$$
