Derive the equation of the tangent line to F(x) at x=1

[phawksbball24]

The problem gives you F(x) = Integral from 0 to 2x of (te^t^2 + cos(pi t)) dt

I am suposed to derive the equation to the tangent line of F(x) at x=1.

(I apologize about some of the notation I don't have any software to write more clearly)

I think the equation I should use to find my slope would be te^t^2 + cos(pi t) because if I were to integrate it first I would then have to turn around and find the derivative since I am determining the tangent line. When I do this I plugged in 1 for t (I am not exactly sure if I can do that since the x is actually at the top of the integral and not in the function itself) and this resulted in a value of e - 1

If I use the point (1, e - 1) and m = e - 1 when I go to find the line it will result in e - 1 = (e - 1)(1)+b and obviously then b = 0, but when I plug both equations into my calculator to check my answer it clearly is not the tangent line. Any assistance would be greatly appreciated, thanks!

 

[royhaas]

You did not calculate the derivative of the integral properly because of the 2x argument. Use the chain rule.

 

[pka]

Your problem properly formatted is \(\displaystyle F(x) = \int\limits_0^{2x} {\left( {te^{t^2 } + \cos (\pi t)} \right)dt}\).

Note that \(\displaystyle F(1) = \frac{e^4}{2}\) , \(\displaystyle F'(x) = 2\left[ {\left( {2x} \right)e^{4x^2 } + \cos (2\pi x)} \right]\) , and \(\displaystyle F'(1) = 4e^4 + 2\).

 

[phawksbball24]

Thank you that helped a lot