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I'm not absolutely sure on how I can deal with this problem with this problem:

Find $ \dfrac{dy}{dx} $ if $ y = 2u^2 - 3u $ and $ u = 4x - 1 $

I am trying to use the chain rule on it.. $ \dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} $

My work so far: $ \dfrac{d}{du}(2u^2-3u) * \dfrac{d}{dx}(4x-1) = (4u-3)(4) $

However I am not absolutely sure I am doing it right.. and I don't have the answer in my book.

Thanks for help, it's appreciated !

EDIT: typos.

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    Note that $\frac{dy}{dx}$ should be given in terms of $x$ and not in terms of $u$. If only there was some way of seeing $u$ as a function of $x$......2012-03-11

3 Answers 3

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What you did is correct, so the final step of yours

$ (4u-3)(4) = 4(4(4x-1)-3) = 16(4x-1)-12 = 64x-16-12 = 64x-28$

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You are correct. But, then you should substitute $u=4x-1$ back in at the end to get

$ 4(4x-1)(4). $

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You can also simplify $ y = 2u^2 - 3u = 2(4x-1)^2-3(4x-1) = 32x^2-28x+5$ Then $\frac{dy}{dx} = 64 x - 28$

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    @Nex: Expansion is only practical if the exponents are relatively small. Imagine trying it for $y=u^{10} +1$.2012-03-11