equation of tangent line of an integral

[quixotiK]

If f(x) = Integral(bottom number 1 top number x) e^(x^2)*-x) then equation of the tangent line to the graph of f at x = 1 is?

This is what i did, its not right though, i subed 1 into e^(x^2)*-x to get
f ' (x) = -2.718....
then i used y = mx+b = -2.718x + b since its from 1 to 1 i assumed that the value of f(1) = 0 thus b = positive 2.718 so the equation would be y = -2.718x + 2.718.

It counted it wrong. Please respond quickly because the assignment is due at 12 and this question could be the difference between an a and a b in the course. Apreciate it, Cody

 

[daon]

If f(x) = Integral(bottom number 1 top number x) e^(x^2)*-x) then equation of the tangent line to the graph of f at x = 1 is?

This is what i did, its not right though, i subed 1 into e^(x^2)*-x to get
f ' (x) = -2.718....
then i used y = mx+b = -2.718x + b since its from 1 to 1 i assumed that the value of f(1) = 0 thus b = positive 2.718 so the equation would be y = -2.718x + 2.718.

It counted it wrong. Please respond quickly because the assignment is due at 12 and this question could be the difference between an a and a b in the course. Apreciate it, Cody

f(x) = e^(x^2) * (-x)
f'(x) = [e^(x^2)]'(-x) + e^(x^2)[-x]'

The derivative of e^(x^2) is found by taking the LN first:
if y = e^(x^2) then ln(y) = ln(e^(x^2))
this implies that ln(y) = x^2. Now take the derivative of both sides implicitly.
(y')/(y) = 2x
so, y' = y*2x, but y = e^(x^2), so, y' = (e^(x^2))*(2x). Whew...

So, f'(x) = ((e^(x^2))*(2x))(-x) - e^(x^2) = -e^(x^2)[2x^2 + 1]

Not sure how you got -e for the derivative...

Anyway, to find the tangent line to the curve, plug 1 into f'(x). That will be the slope of the line at 1, then I think you can handle it from here.

 

[quixotiK]

f(x) does not equal e^(x^2)-x it = the integral of e^(x^2)-x... bottom number on integral is 1 top number is x.

 

[tkhunny]

It's a nasty definition. You cannot have the same variable for the argument and the limit. That makes no sense.

\(\displaystyle \L\,f(x) = \int_{1}^{x}{e^{t^{2}}-t}\,dt\)

\(\displaystyle \L\,f'(x) = \frac{d}{dx}\left({\int_{1}^{x}{e^{t^{2}}-t}\,dt}\right)\,=\,e^{x^{2}}-x\)

The '1' in the lower bound doesn't do anthing, since the integral portion produces a constant value with a subsequent derivative of zero (0).