MHB Solve Derivative of Integral with F(x) - SnowPatrol Yahoo Answers

[MarkFL]

Here is the question:

Derivative of integral?


F(x) = integral of e^(t^2)dt (upper limit = cosx, lower limit = sinx)
Now find F'(x) at x=0

HOW DO I SOLVE THIS :O

Okay so this is what I did,
solve integration, answer is [e^(t^2)]/2t
[2t is the derivative of power(t^2), you're suppose to DIVIDE the integration by the derivative, riiight?]

I put t=cosx, t=sinx
So, it becomes

e^({cosx}^2)]/2cosx - e^({sinx}^2)]/2cosx

Now I take its derivative...
which turns out to be very complicated so I think I'm doing it wrong, because it is supposed to be not-so-long.

THIS IS THE ANSWER GIVEN AT THE BACK:
Answer:
F ′(x)=exp (cos2 (x)) ·−sin (x)−exp (sin2 (x)) · cos (x) by FTOC
F ′(0)=exp (1) · 0−exp (0) · 1=−1

NOW HOLD ON A SECOND.
ISNT DERIVATIVE OF AN INTEGRAL, THE FUNCTION ITSELF? YESSSS.

OKAY, BUT THE DERIVATIVE AND INTEGRAL DONT UMMM CANCEL OUT TILL THE dx/dy/dt IS SAME WITH DERIVATIVE AND INTEGRATION!

Okay, I somehow solved the question B) *pat pat*
Can someone tell me how do i write this down on my paper??

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello SnowPatrol,

We are given:

$$F(x)=\int_{\sin(x)}^{\cos(x)} e^{t^2}\,dt$$

And we are asked to find $F'(0)$.

By the FTOC and the Chain Rule, we have:

$$\frac{d}{dx}\int_{g(x)}^{h(x)} f(t)\,dt=f(h(x))\frac{dh}{dx}-f(g(x))\frac{dg}{dx}$$

Applying this formula, we find

$$F'(x)=-\sin(x)e^{\cos^2(x)}-\cos(x)e^{\sin^2(x)}$$

Hence:

$$F'(0)=-\sin(0)e^{\cos^2(0)}-\cos(0)e^{\sin^2(0)}=0-1=-1$$