Question

what is the equation of the tangent line to y= e^4x + tan-1 (x) at x=0

Answer 1

tangent line = derivative at the point for slope, then just use the point. So derivative of this is 4e^(4x)+1/(x^2+1). At x=0 it is 4+1=5. The point is then (0,0). So use y-y1=m(x-x1) to get y-0=5(x-0), or y=5.

Answer 2

the point is (0,1) that we are trying to find the tangent line

Answer 3

Point is actually (0,1), so line is y-1=5(x-0)

Answer 4

gj mark

Answer 5

I didn't know what the y point was how do you find it?

Answer 6

Plug in x=0 into the original equation

Answer 7

So is e^4x 1 then? Thats what I'm confused about

Answer 8

e^(0)=1

Answer 9

just like 5^(0)=1
and 2^(0)=1

Answer 10

Ooh ok thanks

Answer 11

if tan^(-1)(0)=s
then tan(s)=0
what value of s betweeen -pi/2 and pi/2 makes this expressions true

Answer 12

Ok got it now... thanks so much

Answer 13

np

Answer 14

when i said expression i mean equation lol

Answer 15

yea gotcha