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    Given that y = (secx+tanx)^2

    show that dy/dx = 2 y secx

    i have managed to do this bit

    find d^2y/dx^2 , in terms of y, secx and tanx - is this bit correct?

    d^2y/dx^2 = 4 (secx + tanx) (secxtanx + tanx) secxtanx

    show that cosx(d^2y/dx^2) - 2(dy/dx) = 2 y tanx

    so....

    4 cosxsecxtanx (tanxsecx + secx)(secx +tanx) - 2(secx +tanx)^2 secx= LHS

    where do i go from here???
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    (Original post by Custardcream000)
    find d^2y/dx^2 , in temrs of y =secx and tanx - is this bit correct?
    Please post the question as it was written.
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    (Original post by Custardcream000)
    Given that y = (secx+tanx)^2

    show that dy/dx = 2 y secx

    i have managed to do this bit

    find d^2y/dx^2 , in temrs of y =secx and tanx - is this bit correct?

    d^2y/dx^2 = 4 (secx + tanx) (secxtanx + tanx) secxtanx

    show that cosx(d^2y/dx^2) - 2(dy/dx) = 2 y tanx

    so....

    4 cosxsecxtanx (tanxsecx + secx)(secx +tanx) - 2(secx +tanx)^2 secx= LHS

    where do i go from here???
    Looks absolutely hideous - keep it in terms of y as required by the question, please. Your second derivative is wrong.

    So we know \frac{dy}{dx}=2y\sec(x)

    Therefore by the chain and product rules we get \frac{d^2y}{dx^2}=2\frac{dy}{dx}  \sec(x)+2y\sec(x)\tan(x) and you can sub for dy/dx and keep it in terms of y.
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    so for iii)

    LHS = cosx(2ysecxtanx + 2(dy/dx)secx) -2(2ysecx)
    = 2ysecx (tanxcosx + secx - 2)

    if you cnace whats in the bracket = sinx -2

    so you would get 2ytanx put you would be left with -2
 
 
 
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