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

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

Mathematics
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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.
the point is (0,1) that we are trying to find the tangent line
Point is actually (0,1), so line is y-1=5(x-0)

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Other answers:

gj mark
I didn't know what the y point was how do you find it?
Plug in x=0 into the original equation
So is e^4x 1 then? Thats what I'm confused about
e^(0)=1
just like 5^(0)=1 and 2^(0)=1
Ooh ok thanks
if tan^(-1)(0)=s then tan(s)=0 what value of s betweeen -pi/2 and pi/2 makes this expressions true
Ok got it now... thanks so much
np
when i said expression i mean equation lol
yea gotcha

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