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bettyboop8904
Group Title
find an equation of the tangent line to curve at the given point:
y=e^(2x)cos(pi)x (0,1)
 one year ago
 one year ago
bettyboop8904 Group Title
find an equation of the tangent line to curve at the given point: y=e^(2x)cos(pi)x (0,1)
 one year ago
 one year ago

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bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
\[y=e ^{2x}\cos \pi x\](0,1)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[y'=2e^{2x}\cos(\pi x)\pi e^{2x}\sin(\pi x)\] by the product and chain rule replace \(x\) by \(0\) and find your slope
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
why do you differentiate again?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
you pretty much instantly get \(m=2\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\((fg)'=f'g+g'f\) you have a product, you need the product rule
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\(f(x)=e^{2x},f'(x)=2e^{2x}\) by the chain rule \[g(x)=\cos(\pi x), g'(x)=\pi \sin(\pi x)\] again by the chain rule
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
no no i understand that but you're looking for the slope so what connection does the equation and the derivative have? The slope is connected nvm i answered my own question lol is that what you do every time you want to find an equation of a tangent line like this? is you take the derivative?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
yeah you need the slope and a point
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
slope is the value of the derivative evaluated at the \(x\) value , in your case 0
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
ok so from the beginning again... you said you take the derivative of the function and then plug 0 in for x?
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
@satellite73
 one year ago

RyanL. Group TitleBest ResponseYou've already chosen the best response.1
Basically the function is curved and at the given point we want to know what it looks like if we zoom in on the graph. As we zoom in to that point the graph starts to look like a straight line with a slope. In order to find that slope we find the derivative of the function and plug the x value of the point we zoom in. The output is out slope. Once we have that we can use point slope formula to find the equation.
 one year ago

sweet1137 Group TitleBest ResponseYou've already chosen the best response.1
the tangent line to a curve at a point represents the rate of change of that function at that point. The tangent line not only passes through that point but extends outward with a slope that is equal to the rate of change of the function at that point
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
ok so once i have the slope do i plug it into \[yy _{1}=m(xx _{1})\]
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
with the (0,1)
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
so the answer should be \[y=2x+1\] right? = )
 one year ago

sweet1137 Group TitleBest ResponseYou've already chosen the best response.1
yes, if the value of the derivative at (0,1) is 2
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
woohoo! I just needed a friendly reminder lol = )
 one year ago
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