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amorfide

  • 3 years ago

y=3(x+2)² differentiate it please, but do it via substitution method, i can differentiate this but i can not do the substitution method

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  1. anonymous
    • 3 years ago
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    the derivative of something squared is two times something, times the deriavtive of something

  2. amorfide
    • 3 years ago
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    LOL i know how to do it i just cant do the whole let u= ...

  3. anonymous
    • 3 years ago
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    or you can multiply out an get \[3(x^2+4x+4)=3x^2+12x+12\]\]

  4. amorfide
    • 3 years ago
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    i know i multiply by the power take one off then multiply by derivative of the inside of bracket but i need the substitution method

  5. precal
    • 3 years ago
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    use chain rule

  6. anonymous
    • 3 years ago
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    then ignore it, it is a crutch. if you know the chain rule it is entirely unnecessary to say \(u=x+2\) \(y=u^2\) etc

  7. precal
    • 3 years ago
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    yes, but I believe those were the given instructions

  8. amorfide
    • 3 years ago
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    i really would love the substitution method explained

  9. anonymous
    • 3 years ago
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    i guess you would say \[u=x+2\] \[y=3u^2\] etc etc, but it is silly

  10. amorfide
    • 3 years ago
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    ahhhh okay, then dy/du=6u then du/dx=1 mutliply then substitute u thank you <3

  11. anonymous
    • 3 years ago
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    you have a composition of functions 1) first add 2 2) then square the result, multiply by 3 inner function can be written as \(u=x+2\) outer as \(y=3u^2\)

  12. amorfide
    • 3 years ago
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    so i substitute u= then y is equal to the new equation with u substituted thank you!

  13. anonymous
    • 3 years ago
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    yes, that is correct you get \[6u=6(x+2)\]

  14. precal
    • 3 years ago
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    do you know the chain rule?

  15. precal
    • 3 years ago
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    Take the derivative of the inside, take the derivative of the outside leave the inside alone

  16. precal
    • 3 years ago
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    |dw:1349725718613:dw|

  17. amorfide
    • 3 years ago
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    i know that wy i was struggling with the substitution method i know how to do it straight away though

  18. precal
    • 3 years ago
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    technically the sub method is the chain rule..

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