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Core 3 AQA help?

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    • Thread Starter

    Looked through various sites for a solution to no avail, i'm stuck on Jan 2010 question 7 part b.

    "show that d^2y/dx^2=qy(1+y^2), where q is a number to be determined"

    the previous question found that dy/dx= 4(1+tan^2 4x)

    any help is greatly appreciated!

    Differentiate \frac{dy}{dx} again to get \frac{d^2y}{dx^2}.

    \frac{d}{dx}(4+4\tan^2(4x)) = 0+ \frac{d}{dx}(4\tan^2(4x)) = 4 * \frac{d}{dx}(\tan(4x)\tan(4x))

    Using the product rule:
    4(\tan(4x)*4\sec^2(4x)+\tan(4x)*  4\sec^2(4x)) = 32\tan(4x)\sec^2(4x) = 32y(1+y^2) using the fact that y=\tan(4x) and 1+\tan^2(4x)=\sec^2(4x)

    Chain rule
     \tan^2 4x \equiv (\tan4 x)^2 ,
     \displaystyle \frac{d}{dx} (\tan^24 x)=2\times 4 \sec^2 4x \times \tan 4x=8\tan 4x\sec^2 4x .
    If it helps you can do that whole u sub, and find du/dx but I haven't done that here.
    [I haven't seen the question and didn't realise that you had to find second derivative in terms of y but I hope this helps anyway.]
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