My Math Forum Can anyone explain the specific steps involved

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September 7th, 2016, 07:40 AM   #1
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Can anyone explain the specific steps involved

I know the problem itself involves calculus, but I'm confused about the algebra that is involved here. Can someone specifically explain (step by step) how they got from step 2 to step 3?

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 September 7th, 2016, 07:47 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics $\displaystyle (x-3)^2(1) + (x-2)[2(x-3)] = (x-3)^2 + 2(x-2)(x-3) = (x-3)[(x-3) - 2(x-2)] = (x-3)(x - 3 - 2x - 4)$
September 7th, 2016, 08:30 AM   #3
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Thanks for the response. How did you arrive at this please:
Quote:
 Originally Posted by 123qwerty $\displaystyle (x-3)^2 + 2(x-2)(x-3) = (x-3)[(x-3) - 2(x-2)]$

 September 7th, 2016, 08:55 AM #4 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions The solution presented is the application of the product rule, which can be applied when you're trying to differentiate a function which is formed by the product of smaller functions (or 'bits' or 'chunks' as I like to call them; definitely not official terminology!). In general, let's say that we have $\displaystyle u = u(x)$ and $\displaystyle v = v(x)$, the derivative of $\displaystyle uv$ (i.e. the product of the two functions) with respect to $\displaystyle x$ is: $\displaystyle \frac{d(uv)}{dx} = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}$ That's about as specific as it gets, but if the above looks like jibberish to you, try following these instructions: 1. Take the product you start with (uv) 2. Differentiate first bit 3. Replace that bit in the product with the derivative you just calculated 4. Keep the result to one side 5. Goto step 1, but instead differentiate the next bit 6. Finally: add up all the results together. Let's apply this to your problem 1. Take the product you start with: $\displaystyle uv = (x-2)(x-3)^2$ We have two "bits", u and v. 2. Differentiate the first bit (which is u): $\displaystyle \frac{du}{dx} = \frac{d}{dx}\left(x-2\right) = 1$ 3. Replace that bit in the product with the derivative you just calculated That is, we replace the $\displaystyle (x-2)$ with 1: $\displaystyle \frac{du}{dx} \cdot v = 1 \cdot (x-3)^2 = (x-3)^2$ 4. Keep that result to one side. Okay... go back to 1, but instead work on the next bit So... step 1: take your product $\displaystyle uv = (x-2)(x-3)^2$ 2. Differentiate the next (second) bit (which is v) $\displaystyle \frac{dv}{dx} = \frac{d}{dx}\left((x-3)^2\right) = 2(x-3)$ 3. Replace that bit in the product with the derivative you just calculated That is, we replace the $\displaystyle (x-3)^2$ with $\displaystyle 2(x-3)$: $\displaystyle u \cdot \frac{dv}{dx} = (x-2) \cdot 2(x-3) = 2(x-2)(x-3)$ 4. Keep that result to one side Since that second bit is the last one, we can take our results, which are $\displaystyle (x-3)^2$ and $\displaystyle 2(x-2)(x-3)$ and add them together to finish: $\displaystyle \frac{d(uv)}{dx} = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} = (x-3)^2 + 2(x-2)(x-3)$ This is exactly the same as the line labelled step 2 in the attached answer. The final step is to factorise a factor (x-3) out of both terms and simplify the result, which you should find very easy if you're okay with differentiation. If your factorisation is tricky, then go back and practise a few old factorisation questions. I went into detail about those particular instructions. But... why go to the effort of explaining those instructions? Well... once you get bored with the product rule for two bits, you can use the exact same instructions to calculate the product rule for any number of bits. Here's an example with 3 bits: $\displaystyle f(x) = (x-2)(x^2-3)(x^3-4)$ Let $\displaystyle u = x-2, v = x^2 - 3$ and $\displaystyle w = x^3 - 4$ so that $\displaystyle f(x) = uvw$. The derivative of $\displaystyle f$ with respect to $\displaystyle x$ is is: $\displaystyle \frac{df}{dx} = \frac{du}{dx} \cdot vw + u \cdot \frac{dv}{dx} \cdot w + uv \cdot \frac{dw}{dx}$ $\displaystyle \frac{df}{dx} = 1 \cdot (x^2-3)(x^3-4) + (x-2) \cdot 2x \cdot (x^3 - 4) + (x-2)(x^2-3) \cdot (3x^2)$ Clean it up a bit: $\displaystyle \frac{df}{dx} = (x^2-3)(x^3-4) + 2x(x-2)(x^3 - 4) + 3x^2(x-2)(x^2-3)$
September 7th, 2016, 11:56 AM   #5
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Thank you for taking the time for such an elaborate explanation.

The differentiation I understand, no problem. It's this step here, involving algebra, that I still don't understand. I don't get the factoring here.

Quote:
 Originally Posted by 123qwerty (x−3)^2+2(x−2)(x−3) =(x−3)[(x−3)−2(x−2)]
Shouldn't it be (x-3)^2 + 2(x-2)(x-3), which factors to (x-3)[(x-3)+2(x-2)]? He has a -2(x-2), isn't it supposed to be +2(x-2)???

Last edited by mickapoo; September 7th, 2016 at 12:02 PM.

September 7th, 2016, 12:05 PM   #6
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Quote:
 Originally Posted by 123qwerty $\displaystyle (x-3)^2(1) + (x-2)[2(x-3)] = (x-3)^2 + 2(x-2)(x-3) = (x-3)[(x-3) - 2(x-2)] = (x-3)(x - 3 - 2x - 4)$
Why do you have (x-3)[(x-3) - 2(x-2)], why isn't it (x-3)[(x-3) + 2(x-2)]?

September 7th, 2016, 01:37 PM   #7
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Quote:
 Originally Posted by mickapoo Why do you have (x-3)[(x-3) - 2(x-2)], why isn't it (x-3)[(x-3) + 2(x-2)]?
Hi mickapoo,

It IS +2(x-2).

There's a missing step between step 2 and 3 in your linked solution. It should be:
(x-3)² + (x-2)(2(x-3))
(x-3)² + 2(x-2)(x-3)

Now, we factorize the term (x-3), and get:
(x-3) [(x-3) + 2(x-2)]

We can now distribute the 2 in front of the term +2(x-2), and get:
(x-3) [(x-3)+(2x-4)]

Notice there's a + sign in front of the term +(2x-4), not a multiplication sign, so we actually have too many brackets in that expression. We can clean it up and get:
(x-3)(x-3+2x-4)

We can now add up the similar terms inside the second bracket and get:
(x-3)(3x-7)

Last edited by sKebess; September 7th, 2016 at 01:45 PM.

September 7th, 2016, 07:19 PM   #8
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Quote:
 Originally Posted by mickapoo Thank you for taking the time for such an elaborate explanation. The differentiation I understand, no problem. It's this step here, involving algebra, that I still don't understand. I don't get the factoring here. Shouldn't it be (x-3)^2 + 2(x-2)(x-3), which factors to (x-3)[(x-3)+2(x-2)]? He has a -2(x-2), isn't it supposed to be +2(x-2)???
Ah yes, sorry about the error. I blame it on midnight forum-surfing again

 September 7th, 2016, 10:15 PM #9 Newbie   Joined: Sep 2016 From: Chipping Norton Posts: 5 Thanks: 0 $\dfrac{d}{dx}{\large[}(x - 2)(x - 3)^2\large{]}$ let $u = x - 2\hspace{1in}$ let $v =(x - 3)^2$ $\hspace{20px}u' = 1\hspace{150px} v'= 2(x - 3)$ \begin{align*}y' = u'v + uv' &= (x - 3)^2 + 2(x - 3)(x - 2) \\ &= (x - 3){\large[}(x - 3) + 2(x - 2){\large]} \\ &= (x - 3){\large[}(x - 3) + (2x - 4){\large]} \\ &= (x - 3){\large[}x - 3 + 2x - 4{\large]} \\ &= (x - 3)(3x - 7)\end{align*} Dunno whether this will help... Last edited by skipjack; December 13th, 2016 at 10:29 PM.
 September 15th, 2016, 12:51 AM #10 Newbie   Joined: Jun 2016 From: india Posts: 24 Thanks: 4 Step 2-$\displaystyle (x-3)^2(1)+[(x-2)2(x-3)]$ $\displaystyle \Rightarrow (x-3)^2+[2x-4(x-3)]$ Take (x-3) as a common factor $\displaystyle (x-3)[x-3+2x-4]$ which is step 3

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