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    The curve has the equation, 'y= e^2x tanx'.
    I am asked to show the turning points on this curve occur where
    'tanx = 1'

    I differentiated the equation and got:
    2e^2x . tanx + e^2x sec^2 x = 0

    then I took 'e^2x' out so that I have:

    e^2x (2tanx + sec^2 x) = 0

    afterwards, I changed 'sec^2 x' into '1+tan^2 x' so....

    e^2x (2tanx + 1+tan^2 x) = 0

    I sense that I need to factorise it, but how do the get rid of the
    'e^2x' outside the brackets
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    (Original post by W.H.T)
    The curve has the equation, 'y= e^2x tanx'.
    I am asked to show the turning points on this curve occur where
    'tanx = 1'
    this means that when you equate dy/dx to 0, you need to simplify it until you get tanx - 1 = 0

    so you can just divide by the e^2x
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    e^2x=0

    or
    2 tanx+1+tan^2x=0

    I think. Then factorise the tans.
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    (Original post by W.H.T)

    e^2x (2tanx + 1+tan^2 x) = 0

    I sense that I need to factorise it, but how do the get rid of the
    'e^2x' outside the brackets
    divide 0 by e^2x. Simples
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    (Original post by TylerDurden)
    e^2x=0

    or
    2 tanx+1+tan^2x=0

    I think. Then factorise the tans.
    noting of course that e^2x cannot equal 0
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    Divide both sides by e2x
    (tanx+1)2=0
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    e^{2x} can not equal 0.
    Therefore, simply factorise tan^2x+2tanx+1
    and you would therefore get: (tanx+1)^2
 
 
 
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