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.
$\endgroup$ 13 Answers
$\begingroup$What you did is correct, so the final step of yours
$$ (4u-3)(4) = 4(4(4x-1)-3) = 16(4x-1)-12 = 64x-16-12 = 64x-28$$
$\endgroup$ $\begingroup$You are correct. But, then you should substitute $u=4x-1$ back in at the end to get
$$ 4(4x-1)(4). $$
$\endgroup$ $\begingroup$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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