Question

Find an equation of the tangent line to the curve at the given point.
y = x + tan(x) at (pi, pi)

Answer 1

@exaggerate did you try differentiating?

Answer 2

yeah i got y' = 1 + sec^2(x) and when i plug in pi for x i don't know what 1 + sec^2(pi) is

Answer 3

That's easy
\[\sec \pi=\frac{1}{\cos\pi}\]
\[cos \pi=??\]

Answer 4

-sin but what about the ^2 on the secant

Answer 5

\[\cos \pi= -1\]
We'll use the square
\[1+\sec^2 \pi=1+(-1)^2\]

Answer 6

thanks i think i can get the slope now

Answer 7

Did you understand?

Answer 8

i do now i just didn't know secx = 1/cosx