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Math Help - Question on derivatives of trig functions

  1. #1
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    Question on derivatives of trig functions

    I have a maths B (Grade 11 & 12 Advanced) exam on monday and im studying up for my whole weekend. D=
    Anyway one question that I need help for is:

    y = tan^2 3x

    The answer I got in the end after using the quotient rule was

    dy/dx = 6cos3x * cos^2 3x + sin^2 3x * 6sin 3x / cos^3 3x

    The denominator is correct but the answer says that the numerator is 6sin3x

    Can you please help me using tanx = sinx / cosx not sec^2 x

    Thanks to anyone who can help me =)
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  2. #2
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    Hello, nozpoz!

    You dropped some exponents . . .


    Differentiate with the Quotient Rule: . y \:=\:\tan^23x

    We have: . y \;=\;\frac{\sin^23x}{\cos^23x}


    \text{Then: }\;\frac{dy}{dx} \;=\;\frac{\cos^23x\cdot \bigg[2\cdot\sin3x\cdot\cos3x\cdot 3\bigg] - \sin^23x\cdot\bigg[2\cdot\cos3x(-\sin3x)\cdot 3\bigg]}{\cos^43x}

    . . . . . . . =\;\frac{6\sin3x\cos^33x + 6\sin^33x\cos3x}{\cos^43x}

    . . . . . . . =\;\frac{6\sin3x\cos3x\cdot\overbrace{(\cos^23x + \sin^23x)}^{\text{This is 1}}}{\cos^43x}

    . . . . . . . =\;\frac{6\sin3x}{\cos^33x}


    which is equal to: . 6\sec^23x\tan3x

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