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onegirl
 one year ago
Find an equation of the tangent line at the given value of x.
onegirl
 one year ago
Find an equation of the tangent line at the given value of x.

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tkhunny
 one year ago
Best ResponseYou've already chosen the best response.2What'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?

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.2* "=" in the parentheses should be "". F(x)  F(0)

tkhunny
 one year ago
Best 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.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.2Well, what do you do with a derivative when you want the slope of a tangent line?

agent0smith
 one year ago
Best 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 ?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1We 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.

agent0smith
 one year ago
Best 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 \[\large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}+\pi^2})\;dt = \sin(\sqrt{x^{2}+\pi^2})\]

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.2Ah, well that's silly. Why would I do that? Good call!

agent0smith
 one year ago
Best 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

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Looks like m=0 too, since \[\large m = \sin(\sqrt{0^{2}+\pi^2}) = \sin \sqrt {\pi^2} = \sin \pi = 0\] so y=0
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