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# Help with chain-rule question? y=[(x+3)^4]-4x watch

1. How do I differentiate y=[(x+3)^4]-4x
I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?
2. .

The differential operator is linear, the derivative of a sum of function is the sum of the derivatives of each individual function.
3. (Original post by yoshimatey)
How do I differentiate y=[(x+3)^4]-4x
I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?
This video should explain everything:
http://www.examsolutions.net/maths-r...tutorial-1.php
4. (Original post by yoshimatey)
How do I differentiate y=[(x+3)^4]-4x
I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?
the -4x is a separate term
5. (Original post by yoshimatey)
How do I differentiate y=[(x+3)^4]-4x
I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?
The chain rule, (d/Dx)((x+3)^4)=(du^4)/du) (du/dx) where u=x+3 and (d/du) u^4= 4u^3

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6. just differentiate each bit separately ?
7. (Original post by Zacken)
.

The differential operator is linear, the derivative of a sum of function is the sum of the derivatives of each individual function.
THANKS A BILLION! You have really helped me out!
8. (Original post by yoshimatey)
THANKS A BILLION! You have really helped me out!
You're very welcome!

You can even check this work for certain functions , so

But also,
9. (Original post by Zacken)
You're very welcome!

You can even check this work for certain functions , so

But also,
Or prove it.
10. (Original post by morgan8002)
Or prove it.
Indeed, although it might be a little soon to ask the person to prove it given their unfamiliarity with limits, but for anybody interested:

Let , then

and a bit of shuffling and the sum rule for limits yields

11. (Original post by Zacken)
Indeed, although it might be a little soon to ask the person to prove it given their unfamiliarity with limits, but for anybody interested:

Let , then

and a bit of shuffling and the sum rule for limits yields

'overkill mate' - Ronald Weasley

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Updated: October 25, 2015
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