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    what is the derivative of tan2x ?
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    2sec^2(2x)
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    let u=2x
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    express y = tan2x as y = sin2x / cos2x and use quotient rule.

    You can let u = 2x and work without 2x if it will make the life easier for you. But don't forget it in the final answer.
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    Re: what is the derivative of tan2x ?
    express y = tan2x as y = sin2x / cos2x and use quotient rule.

    You can let u = 2x and work without 2x if it will make the life easier for you. But don't forget it in the final answer.
    I used that method and I got (2(cos2x)^2-2(sin2x)^2)/(cos2x)^2, The actual question in the book is to find dy/dt of the parametric equations t=tanx and y=tan2x when x= 30 degrees, my final gradient was -3 but the book showed it to be 6 i thought my error was in the differentiation of tan2x
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    (Original post by liam251)
    I used that method and I got (2(cos2x)^2-2(sin2x)^2)/(cos2x)^2, The actual question in the book is to find dy/dt of the parametric equations t=tanx and y=tan2x when x= 30 degrees, my final gradient was -3 but the book showed it to be 6 i thought my error was in the differentiation of tan2x
    The - in the numerator should be a +.
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    The - in the numerator should be a +.
    oooooooooohhhhhhhhhhhhh thanks I knew there was a silly error in that differentiation of tan2x
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    y=tan f(x) implies that \frac{dy}{dx} =f'(x)sec^2 (f(x))


    So \frac{dy}{dx} = 2sec^2 (2x)

    You can also prove this result by setting tan2x equal to sin2x/cos2x.
 
 
 
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