Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 $x$...... –  Arthur Fischer Mar 11 '12 at 18:42

3 Answers 3

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

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

share|improve this answer

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

share|improve this answer
    
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
1  
@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

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.