Product Rule/chain rule problem

[jonnburton]

Hi,

I wondered if anyone could tell me how to approach this problem:

Differentiate \(\displaystyle y = (6-3x)(2x-1)^5\)

I can see that this must involve using the product rule and the chain rule but I'm not sure how to break it down into steps.

Any advice about where to start with this would be greatly appreciated.

 

[stapel]

I wondered if anyone could tell me how to approach this problem:

Differentiate \(\displaystyle y = (6-3x)(2x-1)^5\)

I can see that this must involve using the product rule and the chain rule but I'm not sure how to break it down into steps.

Start at the beginning (the outside) and work your way to the end (the inside):

The function is, at its most basic (or "outer" view), a product: (6 - 3x) multiplied by (2x - 1)^5. So you'll apply the Product Rule to this product.

The first factor is 6 - 3x; you'll apply the usual rules for differentiating polynomials.

The second factor is (2x - 1)^5. This is, at its most basic, a power: (something)^5. Apply the usual rule to this. Then, using the Chain Rule, differentiate the "inside" of the power by applying the usual rules to the (something) that was inside the power.

If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you!

 

[jonnburton]

Thank you very much Stapel for explaining that. I will come back to this later this evening and post my progress!

 

[michala20]

Hi,

I wondered if anyone could tell me how to approach this problem:

Differentiate \(\displaystyle y = (6-3x)(2x-1)^5\)

I can see that this must involve using the product rule and the chain rule but I'm not sure how to break it down into steps.

Any advice about where to start with this would be greatly appreciated.

See detailed solution here

< link removed >

 

[jonnburton]

I'm not sure that the link you posted is to a correct solution, Michala20 - the answer the book gives is \(\displaystyle (2x-1)^4(63-36x)\)

I've tried to follow the steps laid out by Stapel but have got a bit stuck.

This is what I've done:

\(\displaystyle (6-3x)(2x-1)^5\)



First Factor:

\(\displaystyle (6-3x)\)

\(\displaystyle \frac{dy}{dx} (6-3x) = -3 \)


Second Factor

\(\displaystyle (2x-1)^5\)

using chain rule,

\(\displaystyle \frac{dy}{dx} (2x-1)^5 = 10 (2x-1)^4 \)



Using product rule

\(\displaystyle f(x)= (6-3x)\)

\(\displaystyle g(x) = (2x -1)^5\)

\(\displaystyle f(x) g'(x) + f'(x) g(x) \)

\(\displaystyle (6-3x) * 10(2x-1)^4 -3(2x-1)^5\)


This is as far as I have been able to get. I think I must have made a mistake somewhere as the solution provided in the book does not contain a power of 5. But I have checked and believe I have differentiated all of the terms correctly.

 

[Deleted member 4993]

[FONT=MathJax_Main]Your work is correct and the answer given in the web-site is correct.

([/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Main]10[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]5[/FONT]

= 3(2x-1)⁴*[10*(2-x) - (2x-1)]

= 3(2x-1)⁴*[20 - 10x - 2x+1]

= 9(2x-1)⁴*[7 - 4x]

How about that......

 

[michala20]

I'm not sure that the link you posted is to a correct solution, Michala20 - the answer the book gives is \(\displaystyle (2x-1)^4(63-36x)\)

I've tried to follow the steps laid out by Stapel but have got a bit stuck.

This is what I've done:

\(\displaystyle (6-3x)(2x-1)^5\)



First Factor:

\(\displaystyle (6-3x)\)

\(\displaystyle \frac{dy}{dx} (6-3x) = -3 \)


Second Factor

\(\displaystyle (2x-1)^5\)

using chain rule,

\(\displaystyle \frac{dy}{dx} (2x-1)^5 = 10 (2x-1)^4 \)



Using product rule

\(\displaystyle f(x)= (6-3x)\)

\(\displaystyle g(x) = (2x -1)^5\)

\(\displaystyle f(x) g'(x) + f'(x) g(x) \)

\(\displaystyle (6-3x) * 10(2x-1)^4 -3(2x-1)^5\)


This is as far as I have been able to get. I think I must have made a mistake somewhere as the solution provided in the book does not contain a power of 5. But I have checked and believe I have differentiated all of the terms correctly.

you did tight, and the steps are correct.
if you take out (2x-1)^4 you get (-3(2x-1)+10(6-3x)), solve this and you get the solution from the book. In the steps there is one more thing taking 9 out, but it's not a must.

 

[jonnburton]

Thank you for your replies! It's good to know my steps were correct.

The only thing is I don't follow how you both proceeded to the answer.

I can't see how:

(6-3x)*10(2x-1)^4 -3(2x-1)^5

becomes:

3(2x-1)^4 * [10 * (2x-1) - (2x-1)]

It seems to me that the power of 5 disappears and that the 10(2x-1)^4 becomes 3(2x-1)^4.


And with Michala20's comment, I can't see how it is possible to take out a factor. And again the power of 5 disappears...

I know I must be failing to see/understand something here but I absolutely can't see what that is!

Thanks again for all your help.

 

[stapel]

I can't see how:

(6-3x)*10(2x-1)^4 -3(2x-1)^5

becomes:

3(2x-1)^4 * [10 * (2x-1) - (2x-1)]

It is indeed by factoring:

(6 - 3x) * 10(2x - 1)^4 - 3(2x - 1)^5

(6 - 3x) * 10 * (2x - 1)^4 - 3 * (2x - 1)^5

(6 - 3x) * 10 * (2x - 1)^4 - 3 * (2x - 1)^4 * (2x - 1)

(6 - 3x) * 10 * (2x - 1)^4 - 3 * (2x - 1)^4 * (2x - 1)

(2x - 1)^4 * (6 - 3x) * 10 - (2x - 1)^4 * 3 * (2x - 1)

(2x - 1)^4 * 10(6 - 3x) - (2x - 1)^4 * 3(2x - 1)

(2x - 1)^4 * [10(6 - 3x) - 3(2x - 1)]

Follow the bolded bits, especially where the one blue and the other red factor get taken out front as the one purple bit. See the factoring now?

 

[jonnburton]

Thanks a lot for showing those steps, Stapel. Yes, I see the factoring now... I would never have thought of doing that; maybe with a lot of practice things like this will become clear to me!

Thanks again!

 

[lookagain]

[FONT=MathJax_Main]Your work is correct and the answer given in the web-site is correct.

([/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Main]10[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]5 [/FONT]

= 3(2x-1)⁴*[10*(2-x) - (2x-1)]

= 3(2x-1)⁴*[20 - 10x - 2x+1]

= 9(2x-1)⁴*[7 - 4x]

How about that......

In the first line, the text didn't come out correctly for the exponents.\(\displaystyle \ \ \ \ \ \ \ (6 - 3x)*10(2x - 1)^4 - 3(2x - 1)^5\)

 

[stapel]

I would never have thought of doing that....

Fair warning: If this sort of factoring is cropping up in your homework, you should expect to see an exercise which requires this factoring on the next test!

 

[jonnburton]

Yes, I'm going to have to learn to deal with it! On the plus side I won't be taking any tests (just yet) - I'm just trying to teach myself maths as I felt it was something I really should understand, having found it difficult in school and not taking it this far. I'm working my way through a textbook and expect that this kind of factoring will crop up in the future though!