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Math Help - Chain rule w/ trigonometric function help

  1. #1
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    Chain rule w/ trigonometric function help

    find the derivative:
    \frac{d}{dx}(\cot^4 t -\csc^4 t)

    What did i do wrong?
    4(-\csc^2 t)^3 - 4(-\csc t\cot t)^3
    -4\csc^6 t + 4\csc^3 t\cot^3 t


    Answer at back of book: 4\cot t \csc^2 t


    Thx & happy new year folks
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  2. #2
    Eater of Worlds
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    Hey Adam.

    Chain rule:

    \frac{d}{dt}[cot^{4}(t)]=-4cot^{3}(t)csc^{2}(t)

    \frac{d}{dt}[csc^{4}(t)]=-4csc^{3}(t)(csc(t)cot(t))=-4csc^{4}(t)cot(t)

    {-}4cot^{3}(t)csc^{2}(t)+4csc^{4}(t)cot(t)

    Factor:

    4cot(t)csc^{2}(t)\underbrace{(csc^{2}(t)-cot^{2}(t))}_{\text{equals 1}}

    \boxed{4cot(t)csc^{2}(t)}
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  3. #3
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    Hello, ^_^Engineer_Adam^_^!

    Your chain rule is a bit off . . .


    Find the derivative: . f(x) \:=\:\cot^4\!t -\csc^4\!t

    Answer: . 4\cot t \csc^2\!t

    We have: . f(x)\:=\:(\cot t)^4 - (\csc t)^4

    Then: . f'(t) \:=\:4\cot^3\!t(-\csc^2t) - 4\csc^3\!t(-\csc t\cot t) \:= \:-4\csc^2\!t\cot^3\!t + 4\csc^4\!t\cot t

    \text{Factor: }\;f'(t)\:=\:4\cot t\csc^2\!t\underbrace{(\csc^2\!t - \cot^2\!t)}_{\text{This equals 1}}

    Therefore: . \boxed{f'(x) \:=\:4\cot t\csc^2t}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We could have simplified at the very beginning . . .

    f(t)\:=\:\cot^4t - \csc^4t \:=\:\underbrace{(\cot^2x - \csc^2t)}_{\text{This equals -1}}(\cot^2t + \csc^2t)

    We have: . f(t) \:=\:-\cot^2t - \csc^2t

    Then: . f'(t)\:=\:-2\cot t(-\csc^2t) - 2\csc t(-\csc t\cot t)

    . . . . . f'(t) \:=\:2\cot t\csc^2\!t +2\cot t\csc^2\!t \:=\:\boxed{4\cot t\csc^2\!t}

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