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onegirl

  • 3 years ago

Find the derivative of f(x) = cos x, a = pi/2

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  1. onegirl
    • 3 years ago
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    @zepdrix i ddi it but can u check my answer?

  2. zepdrix
    • 3 years ago
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    sure :O what are you suppose to do? Just find the derivative then plug \(\large a\) in? Or does this have something to do with approximation?

  3. onegirl
    • 3 years ago
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    okay my teacher said there is an equation, so here is what i did f(x) = cos (x) f'(x) = -sin(x) x = a = pi/2 so f'(pi/2) = -sin (pi/2) = -1

  4. zepdrix
    • 3 years ago
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    Your teacher said there is an equation? :) That's not much detail lolol Your derivative looks correct at least.

  5. onegirl
    • 3 years ago
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    yea she said try again there is an equation

  6. onegirl
    • 3 years ago
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    let me show you how i wrote it first okay?

  7. onegirl
    • 3 years ago
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    i mean the answer i wrote first

  8. zepdrix
    • 3 years ago
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    k

  9. onegirl
    • 3 years ago
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    d/dx ( cos (x)) = -sin(x) -sin (pi/2) = -1, so the slope is -1, y = mx + b --> y = =1x + b cos (pi/2) = 0 so the coordinates pare are (pi/2, 0) so y = 0x + b, 0(pi/2) + b equals to 0, 0 = y = 0x + 0 = y = 0 for my final answer.

  10. zepdrix
    • 3 years ago
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    Oh so like the previous problems, you're looking for an equation of the tangent line?

  11. onegirl
    • 3 years ago
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    yes sorry i made a mistake

  12. onegirl
    • 3 years ago
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    yes yes the equation of the tangent line

  13. zepdrix
    • 3 years ago
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    \[\large y=-x+b\] From here you plugged in your point \(\large \left(\pi/2,\;0\right)\). \[\large 0=-\pi/2+b \qquad \qquad \rightarrow \qquad \qquad b=\pi/2\] I think you plugged your \(\large 0\) into the wrong spot.

  14. onegirl
    • 3 years ago
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    ohhh okay

  15. onegirl
    • 3 years ago
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    so instead of 0 = 0(pi/2) + b what would i write

  16. zepdrix
    • 3 years ago
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    \[\huge (\color{royalblue}{\pi/2},\;\color{orangered}{0}) \qquad = \qquad (\color{royalblue}{x},\;\color{orangered}{y})\] Plug them into here, match up the colors.\[\huge \color{orangered}{y}=(-1)\color{royalblue}{x}+b\] I'm not sure why you're plugging a \(\large 0\) in for your -1....

  17. onegirl
    • 3 years ago
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    okay

  18. onegirl
    • 3 years ago
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    so 0 = (-1)(pi/2) + b ?

  19. zepdrix
    • 3 years ago
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    yes

  20. onegirl
    • 3 years ago
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    okay

  21. onegirl
    • 3 years ago
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    thx

  22. zepdrix
    • 3 years ago
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    np

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