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bettyboop8904

  • 2 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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  1. bettyboop8904
    • 2 years ago
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    \[y=e ^{2x}\cos \pi x\](0,1)

  2. satellite73
    • 2 years ago
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    \[y'=2e^{2x}\cos(\pi x)-\pi e^{2x}\sin(\pi x)\] by the product and chain rule replace \(x\) by \(0\) and find your slope

  3. bettyboop8904
    • 2 years ago
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    why do you differentiate again?

  4. satellite73
    • 2 years ago
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    you pretty much instantly get \(m=2\)

  5. satellite73
    • 2 years ago
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    ?

  6. satellite73
    • 2 years ago
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    \((fg)'=f'g+g'f\) you have a product, you need the product rule

  7. satellite73
    • 2 years ago
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    \(f(x)=e^{2x},f'(x)=2e^{2x}\) by the chain rule \[g(x)=\cos(\pi x), g'(x)=-\pi \sin(\pi x)\] again by the chain rule

  8. bettyboop8904
    • 2 years ago
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    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?

  9. satellite73
    • 2 years ago
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    yeah you need the slope and a point

  10. satellite73
    • 2 years ago
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    slope is the value of the derivative evaluated at the \(x\) value , in your case 0

  11. bettyboop8904
    • 2 years ago
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    ok so from the beginning again... you said you take the derivative of the function and then plug 0 in for x?

  12. bettyboop8904
    • 2 years ago
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    @satellite73

  13. RyanL.
    • 2 years ago
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    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.

  14. sweet1137
    • 2 years ago
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    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

  15. bettyboop8904
    • 2 years ago
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    ok so once i have the slope do i plug it into \[y-y _{1}=m(x-x _{1})\]

  16. bettyboop8904
    • 2 years ago
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    with the (0,1)

  17. bettyboop8904
    • 2 years ago
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    so the answer should be \[y=2x+1\] right? = )

  18. sweet1137
    • 2 years ago
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    yes, if the value of the derivative at (0,1) is 2

  19. bettyboop8904
    • 2 years ago
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    woo-hoo! I just needed a friendly reminder lol = )

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