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

Question

I'm not absolutely sure on how I can deal 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 the following: $\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx}.$

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

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

Answer 1

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

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

Answer 2

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

Answer 3

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

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