Simplifying the derivative after using the Product Rule. Help?

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Homework Help Overview

The discussion revolves around simplifying the derivative of a function defined as the product of two polynomial expressions raised to powers. The subject area is calculus, specifically focusing on the application of the product rule for differentiation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the simplification of the derivative expression, discussing common factors and how to factor them out. Some participants express uncertainty about their simplification steps and seek confirmation on their reasoning.

Discussion Status

Several participants have provided guidance on identifying common factors and suggested ways to simplify the derivative. There is an ongoing exploration of the correct application of differentiation rules, with some participants questioning their previous steps and seeking clarification on their understanding.

Contextual Notes

Some participants have noted potential mistakes in their calculations and expressed confusion regarding the proper application of the product rule and chain rule in differentiation. There are also indications of varying interpretations of the original function's notation.

Vanessa13
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Homework Statement


f(x) = (x^3+6)^4 * (x^3+4)^6


Homework Equations


f'(x) = (x^3+6)^4*18x^2(x^3+4)^5 + (x^3+4)^6*12x^2(x^3+6)^3


The Attempt at a Solution


Obviously this expression can be simplified, but I haven't the slightest clue on how to do this.

I would really appreciate the help!
 
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Vanessa13 said:

Homework Statement


f(x) = (x^3+6)^4 * (x^3+4)^6


Homework Equations


f'(x) = (x^3+6)^4*18x^2(x^3+4)^5 + (x^3+4)^6*12x^2(x^3+6)^3


The Attempt at a Solution


Obviously this expression can be simplified, but I haven't the slightest clue on how to do this.

I would really appreciate the help!

Your derivative is made up of two terms:
18x2(x3+4)5(x3+6)4
and
12x2(x3+4)6(x3+6)3

Obviously, both terms have a factor of 6x2. Both terms also have factors of (x3 + 4) and (x3 + 6).

For each of these factors, bring out the smallest power that is present in both terms.

You should end up with 6x2(x3 + 4)m(x3 + 6)n(something + something else)
 
Mark44 said:
Your derivative is made up of two terms:
18x2(x3+4)5(x3+6)4
and
12x2(x3+4)6(x3+6)3

Obviously, both terms have a factor of 6x2. Both terms also have factors of (x3 + 4) and (x3 + 6).

For each of these factors, bring out the smallest power that is present in both terms.

You should end up with 6x2(x3 + 4)m(x3 + 6)n(something + something else)

I'm beginning to understand. If you don't mind, I'm going to show you what I did step by step... Hopefully I do it right.

6x2(x3+4)5(x3+6)3* 3(x3+6) + 12(x3+4) =

6x2(x3+4)5(x3+6)3* (3x3+12+12x3+48) =

6x2(x3+4)5(x3+6)3*(15x3+60)

Did I do the simplifying right?
 
Vanessa13 said:
I'm beginning to understand. If you don't mind, I'm going to show you what I did step by step... Hopefully I do it right.

6x2(x3+4)5(x3+6)3* 3(x3+6) + 12(x3+4) =
You have a mistake in the line above, and that affects everything below here.
6x2(x3+4)5(x3+6)3* [3(x3+6) + [STRIKE]12[/STRIKE]2(x3+4)]

Can you take it from here? There's also another mistake below - 3*6 = 18, not 12.
Vanessa13 said:
6x2(x3+4)5(x3+6)3* (3x3+12+12x3+48) =

6x2(x3+4)5(x3+6)3*(15x3+60)

Did I do the simplifying right?
 
When I was a young student I used to think there was virtue in long formulae like those and if I could handle them it showed I was smart. If anything it showed I wasn't. Took me quite some time to cotton on to the where where where of mathematicians.

I.e. saying instead of your formulae f(x) = 18x2(x3+4)5(x3+6)4 you can say differentiate 18x2u5v4 where u = (x3+4) and v = (x3+6).

Then
f'(x) = 18( 2xu5v4 + 5x2u4u'v4 + 4x2u5v3v') .

You can already see you can take out a factor xu4v3, and further down there may be more simplifications. That way you see what you are doing, the essentials of it. (Of course you do have to remember that the u, v are functions of x so you have to use the chain rule and write not 5u4 but 5u4u' .) Then at the end of the calculation you'll often enough convert back to original variables.

There is no virtue in unnecessarily complicated expressions (and while I am at it, unnecessarily general ones either - you will soon come up across the expression 'without loss of generality'.)
 
Mark44 said:
You have a mistake in the line above, and that affects everything below here.
6x2(x3+4)5(x3+6)3* [3(x3+6) + [STRIKE]12[/STRIKE]2(x3+4)]

Can you take it from here? There's also another mistake below - 3*6 = 18, not 12.

I always make basic errors.

Alright. So with those fixed, this is what my simplified derivative came out to be:

6x2(x3+4)5(x3+6)3(5x3+26)

Better?

Also, THANK YOU!
 
:) Thank you so much!
 
I got confused again :(

f(x) = 7cos2tsin6t

I separated them like this: u= 7cos v =2t y =sin6t
u'= -sin v'= 2 y'= cos6

So far my derivative looks like this:

f'(x) = -sin2tsin6t + 7cos2sin6t + 7cos2tcos6

This is the farthest it can be simplified, right?
 
  • #10
It probably would have been better if you started a new thread with the new problem. :wink:
Vanessa13 said:
I got confused again :(

f(x) = 7cos2tsin6t
Is the function
f(x) = 7 cos(2t) sin(6t)?
 
  • #11
eumyang said:
It probably would have been better if you started a new thread with the new problem. :wink:

Is the function
f(x) = 7 cos(2t) sin(6t)?

That's probably true... Thank you for being my savior! :)

It doesn't have parenthesis around them in my book, so I'm not sure at all... That's why I was so confused.
 
  • #12
eumyang said:
It probably would have been better if you started a new thread with the new problem. :wink:

Is the function
f(x) = 7 cos(2t) sin(6t)?

Okay, I did it this way and it makes a lot more sense.

f'(x)= -sin2tsin6t + 7cos2tcos6t

I think that's all I can do with it, right?
 
  • #13
Vanessa13 said:
Okay, I did it this way and it makes a lot more sense.

f'(x)= -sin2tsin6t + 7cos2tcos6t

I think that's all I can do with it, right?

Simplification aside, that's not the derivative of 7*cos(2t)*sin(6t). Just like the derivative of cos(2t) is not -sin(2t). You have to use the chain rule.
 
  • #14
Dick said:
Simplification aside, that's not the derivative of 7*cos(2t)*sin(6t). Just like the derivative of cos(2t) is not -sin(2t). You have to use the chain rule.

I see that now.

Here goes, third times a charm (hopefully)...

v' = -7sin2sin6t + 7cos2tcos6 + 0 * cos2tsin6t

= -7sin2sin6t + 7cos2tcos6 (then you take out a 7, right?)

v' = 7(-sin2sin6t)(cos2tcos6)

I'm not sure if I'm still supposed to be adding them [(-sin2sin6t) + (cos2tcos6)] or not, but in the example earlier, getting rid of common things also got rid of the addition...

Am I right now?

(Thank you for helping :] )
 
  • #15
Vanessa13 said:
I see that now.

Here goes, third times a charm (hopefully)...

v' = -7sin2sin6t + 7cos2tcos6 + 0 * cos2tsin6t

= -7sin2sin6t + 7cos2tcos6 (then you take out a 7, right?)

v' = 7(-sin2sin6t)(cos2tcos6)

I'm not sure if I'm still supposed to be adding them [(-sin2sin6t) + (cos2tcos6)] or not, but in the example earlier, getting rid of common things also got rid of the addition...

Am I right now?

(Thank you for helping :] )

You are very welcome, but you are writing some serious nonsense here. Start with something simpler. What's the derivative of cos(2t)? Explain your answer. And use parentheses like eumyang suggested.
 
  • #16
I feel really ignorant now.

derivative of cos(2t) = -2sin(2t)

derivative of sin(6t) = 6cos(6t)

derivative of 7 = 0
 
  • #17
Vanessa13 said:
I feel really ignorant now.

derivative of cos(2t) = -2sin(2t)

derivative of sin(6t) = 6cos(6t)

derivative of 7 = 0

That's great! Now use the product rule on 7 cos(2t) sin(6t). There won't be anything like sin(2) in there, will there?
 
  • #18
Dick said:
That's great! Now use the product rule on 7 cos(2t) sin(6t). There won't be anything like sin(2) in there, will there?

v' = -2sin(2t)sin(6t)*7 + cos(2t)*6cos(6t)*7 + 0

v' = -14sin(2t)sin(6t) + 42cos(2t)cos(6t) [Took 14 out...]

v' = 14(-sin(2t)sin(6t)) + 3(cos(2t)cos(6t)) [Do I continue to add them?]
 
  • #19
Vanessa13 said:
v' = -2sin(2t)sin(6t)*7 + cos(2t)*6cos(6t)*7 + 0

v' = -14sin(2t)sin(6t) + 42cos(2t)cos(6t) [Took 14 out...]

v' = 14(-sin(2t)sin(6t)) + 3(cos(2t)cos(6t)) [Do I continue to add them?]

No. I think you should leave it as that. It's about as simple as it's going to get.
 
  • #20
Thank you.

Would you mind if I ever messaged you with a question now and then?

As you can probably tell, Calculus isn't exactly my strong suit.
 
  • #21
Vanessa13 said:
Thank you.

Would you mind if I ever messaged you with a question now and then?

As you can probably tell, Calculus isn't exactly my strong suit.

You can message me if you post the question and you feel you aren't getting the answer you want after a while. Send me link to the post. Don't send the question through a message. Give other people a chance to help. I'm not around 24/7.
 
  • #22
In the original equation you probably meant f(t) not f(x).

Vanessa13 said:
I feel really ignorant now.

Get inside it. It's not an arbitrary rule you are forced to remember and feel stupid if you don't.

Draw any old curve graph and call it f(t) .

Remember what 'function' means and example exercises probably done to familarise you right at the start.

15n14ko.jpg


f(2t) gets to the height that f(t) does in half as much time. It is increasing/decreasing twice as fast.

f'(2t) = 2 f'(t).




( Same idea are f'(kt) = k f'(t)

f'(g(t)) = f'(g(t))g'(t) )
 

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