How do I find an equation of the tangent line to the graph? Find an equation of the tangent line to the graph of $\arctan(x+y)=y^2+\frac{\pi}{4}$ at $(1, 0)$.
This is what I have so far: 
$1 + \frac{dy}{dx}=2y\cdot\sec^2(y^2+\frac{\pi}{4})\cdot\frac{dy}{dx}$
How do I go on from here? 
 A: So from what you have, get $\frac{dy}{dx}$ by itself. If you do that you should have
$$\frac{dy}{dx}=\frac{1}{2y\cdot \sec^2{\left(y^2+\frac{\pi}{4}\right)-1}}$$
Now, there is no need to try and simplify since you are only interested in a tangent line.  Replace all $x$'s with $1$ and all $y$'s with $0$.  Then the slope of the tangent line is
$$\frac{1}{2\cdot0\cdot \sec^2{\left(0^2+\frac{\pi}{4}\right)-1}}=\frac{1}{-1}=-1$$
Thus the equation of the tangent line is 
$$y-0=-1(x-1)\rightarrow y=1-x$$
A: Tangent line is described by first derivative. Basically, you should find y', then find y' at the point and you will have the slope of the tangent line. 
Thus, y' (1,0) is slope of line. Use linear equation for the line, $y_0=kx_0+b$. Subscript 0 is about the point (1,0); k is the slope. Then extrapolate as you know k and b. 
Can you (as you know the slope of line at the point) find equation? If you need furhter help, leave a comment below.
A: $ ( x=1,y=0) $ to be plugged in into the relation you already got by differentiation, to evaluate slope at tangent point.. then and there as $ ( y^{'}=-1)$. Luckily $y=0$ no need to find actual graph by integration.
Equation of tangent (point-slope form) is
$$ \dfrac{y-0}{x-1}=-1 \rightarrow x+y=1. $$
