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

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

satellite73
 3 years ago
Best 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

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0why do you differentiate again?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0you pretty much instantly get \(m=2\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0\((fg)'=f'g+g'f\) you have a product, you need the product rule

satellite73
 3 years ago
Best 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

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0no 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?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0yeah you need the slope and a point

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0slope is the value of the derivative evaluated at the \(x\) value , in your case 0

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

RyanL.
 3 years ago
Best ResponseYou've already chosen the best response.1Basically 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.

sweet1137
 3 years ago
Best ResponseYou've already chosen the best response.1the 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

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

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

sweet1137
 3 years ago
Best ResponseYou've already chosen the best response.1yes, if the value of the derivative at (0,1) is 2

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0woohoo! I just needed a friendly reminder lol = )
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