# Find dy/dx if y = .. and u = ..

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 $x$...... – arjafi Mar 11 '12 at 18:42

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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Wow.. our teacher didnt taught us this way of doing it.. it's really simple and it works for every similar problems I have so far... – Nex Mar 11 '12 at 18:58
@Nex it will only work if substitution will leave you with $y = f(x)$ and no $u$'s. Also, your teacher probably wants you to practice the chain rule. – user2468 Mar 11 '12 at 19:00
@Nex: Expansion is only practical if the exponents are relatively small. Imagine trying it for $y=u^{10} +1$. – Patrick Mar 11 '12 at 20:48

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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