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dy/dx sec^2(3X) watch

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    dy/dx  sec^23x

    where does the 6 come in? (i know it does through answers)

    i got  3sec^23xtan^23x
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    the 6 comes from differentiating sec^2(3x) -----> 2sec(3x) and differentiating 3x ------> 3

    2*3 = 6
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    The 6 comes from the power (2) multiplied by the 3 'inside' the (sec 3x)^2 , by using chain rule

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     \frac{d}{dx} \sec^2 3x

    Let  y = \sec^2 3x

    Let u = \sec 3x

     \Rightarrow y = u^2

     \frac{dy}{du} = 2u

     \frac{du}{dx} = 3\sec 3x \tan 3x

     \Rightarrow \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

     = 2u \times 3\sec 3x \tan 3x

     = 6 \sec^2 3x \tan 3x

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    thats what i was thinking but what about the tan^2x doing something similar?
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    (Original post by picer)
    thats what i was thinking but what about the tan^2x doing something similar?
    use the chain rule with u=sec3x
 
 
 
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Updated: February 13, 2010

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