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tkhunnyBest ResponseYou've already chosen the best response.2
What's your plan? \(\int\limits_{0}^{x}f(x)\;dx = F(x)  F(0)\) where \(F(x)\) is an antiderivative of \(f(x)\). \(\dfrac{d}{dx}(F(x)=F(0)) = F'(x) = f(x)\) Isn't that interesting?
 11 months ago

tkhunnyBest ResponseYou've already chosen the best response.2
* "=" in the parentheses should be "". F(x)  F(0)
 11 months ago

tkhunnyBest ResponseYou've already chosen the best response.2
* Wow! I must be sleeping. I used 'x' two different ways, too! Okay, now that I've confused you. let's just show you. \(\dfrac{d}{dx}\int\limits_{0}^{x}\sin(\sqrt{t^{2}\pi^2})\;dt = \sin(\sqrt{x^{2}\pi^2})\) Ponder on it.
 11 months ago

tkhunnyBest ResponseYou've already chosen the best response.2
Well, what do you do with a derivative when you want the slope of a tangent line?
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
\[\Large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}\pi^2})\;dt = \sin(\sqrt{x^{2}\pi^2})\] You remember how to do that part from earlier, right @onegirl ?
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
We need a tangent line y = mx+b So to find the tangent line at x=0, we need the slope of the line m... we've already differentiated it above, now we need to find m by plugging in x=0. Then we find b by using the original function in the question.
 11 months ago

onegirlBest ResponseYou've already chosen the best response.0
okay i think i got it
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
@tkhunny you put the equation in wrong ( where a + should be), and i copied it wrongly... should be \[\large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}+\pi^2})\;dt = \sin(\sqrt{x^{2}+\pi^2})\]
 11 months ago

tkhunnyBest ResponseYou've already chosen the best response.2
Ah, well that's silly. Why would I do that? Good call!
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
@onegirl did you get the slope m by putting x= 0 into \[\large m = \sin(\sqrt{x^{2}+\pi^2})\] Then find b by using \[\large \dfrac{d}{dx}\int\limits\limits\limits_{0}^{0}\sin(\sqrt{t^{2}+\pi^2})\;dt = 0\] (since the area from 0 to 0 is zero) So since b=0 it'll be y = mx
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Looks like m=0 too, since \[\large m = \sin(\sqrt{0^{2}+\pi^2}) = \sin \sqrt {\pi^2} = \sin \pi = 0\] so y=0
 11 months ago
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