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Quick Help - Differentiating

  1. May 11, 2005 #1
    What are the rules for differentiating tan, sin & cos?

    I know cos = -sin

    tan = sin/cos?
  2. jcsd
  3. May 11, 2005 #2
    Be careful of what you type.

    You can derive sin's derivative (say that five times fast) using the limit definition, from there a simple Taylor series expansion will get you the other two, or knowing sinx' = cosx and cosx' = -sinx, you can use the quotient rule to find tanx.
  4. May 11, 2005 #3
    Yea, just like what Whozum said...

    if [tex]f(x) = \sin \theta[/tex],[tex]f'(x) = \cos \theta[/tex]
    [tex]f(x) = \cos \theta[/tex], [tex]f'(x) = -\sin \theta[/tex]
    [tex]f(x) = \tan \theta[/tex], [tex]f'(x) = \sec^2 \theta[/tex]
    [tex]f(x) = \csc \theta[/tex], [tex]f'(x) = -\csc\theta \cot\theta[/tex]
    [tex]f(x) = \sec \theta[/tex], [tex]f'(x) = \sec\theta \tan\theta[/tex]
    [tex]f(x)= \cot\theta[/tex], [tex]f'(x) = -\csc^2\theta[/tex]

    Try finding the derivative of the inverse trig functions :rolleyes:
  5. May 12, 2005 #4
    You are correct [tex]\frac{d}{dx}\cos{x} = -\sin{x}[/tex]

    ,and I assume you know that [tex]\frac{d}{dx}\sin{x} = \cos{x}[/tex].

    Using these two definitions, use the quotient rule to find the derivative of tanx as Whozum said above.

    [tex]\frac{d}{dx}\frac{u}{v} = \frac{vu' - uv'}{v^2}[/tex]

    Express tangent as [tex]\frac{\sin{x}}{\cos{x}}[/tex] and see what you get using the rule above.

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