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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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\[y=e ^{2x}\cos \pi x\](0,1)
\[y'=2e^{2x}\cos(\pi x)-\pi e^{2x}\sin(\pi x)\] by the product and chain rule replace \(x\) by \(0\) and find your slope
why do you differentiate again?

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Other answers:

you pretty much instantly get \(m=2\)
\((fg)'=f'g+g'f\) you have a product, you need the product rule
\(f(x)=e^{2x},f'(x)=2e^{2x}\) by the chain rule \[g(x)=\cos(\pi x), g'(x)=-\pi \sin(\pi x)\] again by the chain rule
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?
yeah you need the slope and a point
slope is the value of the derivative evaluated at the \(x\) value , in your case 0
ok so from the beginning again... you said you take the derivative of the function and then plug 0 in for x?
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.
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
ok so once i have the slope do i plug it into \[y-y _{1}=m(x-x _{1})\]
with the (0,1)
so the answer should be \[y=2x+1\] right? = )
yes, if the value of the derivative at (0,1) is 2
woo-hoo! I just needed a friendly reminder lol = )

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