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anonymous

  • one year ago

find dy/dx for y = 3cos(x) + csc(x)

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  1. anonymous
    • one year ago
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    Do you know your derivative rules for trig functions? If not, you can find some here: http://www.sosmath.com/calculus/diff/der03/der03.html

  2. anonymous
    • one year ago
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    Heere is also a list of basic rules. This is a simple example where you have two functons added together. http://tutorial.math.lamar.edu/pdf/Common_Derivatives_Integrals.pdf

  3. anonymous
    • one year ago
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    okay so do i just ignore the fact that its dy/dx because the y is already isolated?

  4. anonymous
    • one year ago
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    Can you tell me what \[\Large \frac{ dy }{ dx } 3\cos(x)= ?\]

  5. anonymous
    • one year ago
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    -3sin(x)

  6. anonymous
    • one year ago
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    Ok and what is \[\Large \frac{dy}{dx} \csc(x)? \]

  7. anonymous
    • one year ago
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    -csc(x)cot(x)

  8. anonymous
    • one year ago
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    And when you get that, you add the two together. When you do a derivative of something in the dorm of \[\Large y=f(x)\] You actually have \[\Large \frac{ dy }{ dx }= f'(x)\] This is known as implicit differentiation. You will learn it later on. It is the underlying concept as to what you are doing.

  9. anonymous
    • one year ago
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    So your final answer is -3sin(x)-csc(x)cot(x)

  10. anonymous
    • one year ago
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    okay so then its just -3sin(x)-csc(x)cot(x)

  11. anonymous
    • one year ago
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    It's actually -3sin(x) + (-csc(x)cot(x)) but a positive plus a minus is a minus

  12. anonymous
    • one year ago
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    thank you very much !!

  13. anonymous
    • one year ago
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    Great job! Np

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