anonymous
  • anonymous
what is the equation of the tangent line to y= e^4x + tan-1 (x) at x=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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.
myininaya
  • myininaya
the point is (0,1) that we are trying to find the tangent line
anonymous
  • anonymous
Point is actually (0,1), so line is y-1=5(x-0)

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

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