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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

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

    • one year ago
  2. satellite73 Group Title
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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

    • one year ago
  3. bettyboop8904 Group Title
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    why do you differentiate again?

    • one year ago
  4. satellite73 Group Title
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    you pretty much instantly get \(m=2\)

    • one year ago
  5. satellite73 Group Title
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    ?

    • one year ago
  6. satellite73 Group Title
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    \((fg)'=f'g+g'f\) you have a product, you need the product rule

    • one year ago
  7. satellite73 Group Title
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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

    • one year ago
  8. bettyboop8904 Group Title
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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?

    • one year ago
  9. satellite73 Group Title
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    yeah you need the slope and a point

    • one year ago
  10. satellite73 Group Title
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    slope is the value of the derivative evaluated at the \(x\) value , in your case 0

    • one year ago
  11. bettyboop8904 Group Title
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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?

    • one year ago
  12. bettyboop8904 Group Title
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    @satellite73

    • one year ago
  13. RyanL. Group Title
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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.

    • one year ago
  14. sweet1137 Group Title
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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

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

    • one year ago
  16. bettyboop8904 Group Title
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    with the (0,1)

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

    • one year ago
  18. sweet1137 Group Title
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    yes, if the value of the derivative at (0,1) is 2

    • one year ago
  19. bettyboop8904 Group Title
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    woo-hoo! I just needed a friendly reminder lol = )

    • one year ago
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