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Find derivative with chain rule

  1. Aug 30, 2006 #1
    how would i find the derivative of y= [tex]-18 \sin 80 t[/tex]?
    thanks pavadrin
  2. jcsd
  3. Aug 30, 2006 #2


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    1. Use the fact, which you should already have learned, that the derivative of sin(x) is cos(x).
    2. Use the fact, which you should already have learned, that the derivative of 80t is 80.
    3. Use the chain rule.

    By the way, the derivative is one of the basic operations in calculus so this is clearly not "pre-calculus". I'm moving this to "Calculus and beyond".
  4. Aug 30, 2006 #3
    okay sorry for having placed it in the wrong section, um....ive never differentiated a trig f(x) before thats all and it was the first i came across that i needed to differentiate. could i please given a few hints on the basics? thanks
  5. Aug 30, 2006 #4


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    I would imagine if one is studying differential calculus, that one is using a textbook and within the text book are trigonometric identities and examples.

    HallsofIvy mentioned the chain rule, which is very basic in differentiation.

    Given y = f(g(x)), y' = dy/dx = f'(g(x))*g'(x).

    Has one proved to oneself the definition of a derivative, and then used that definition to find the derivative of various functions?

    y'(x) = dy(x)/dx = [tex]\lim_{\substack{\Delta{x}\rightarrow 0}} \frac{y(x+\Delta{x})-y(x)}{\Delta{x}}[/tex]



    Last edited: Aug 30, 2006
  6. Aug 31, 2006 #5
    okay thanks for the replies and the links. so the derivative of sin (x) = cos (x) therefore that if y = -18sin (80t) then y' = 80(-18cos (80t)), expanding the brackets is equal to y' = -1440cox 80t? thanks
  7. Aug 31, 2006 #6


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    Yup, this is correct. :smile:
    A harmless typo though, cox should read cos, instead. :)
  8. Sep 2, 2006 #7
    oh okay thanks for confirming that :smile:
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