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clara1223

  • one year ago

Calculate the derivative of the given function f(x)=8sin^3(√x) a) f′(x)=24sin^2(√x)cos(√x) b) f′(x)=48sin^2(√x)cos(√x)/x√x c) f′(x)=24cos(√x) d) f′(x)=12sin^2(√x)cos(√x)/√x e) f′(x)=12cos(√x)/√x

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  1. clara1223
    • one year ago
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    yes, one second

  2. Jhannybean
    • one year ago
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    Sorry, I made a typo. \[f(x) = 8\sin^3(x^{1/2})\]

  3. clara1223
    • one year ago
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    actually could you help me with this really quickly? its more pressing: Evaluate (g∘f)′(6), given that f(4)=4, f(5)=6, f(6)=6, g(4)=5, g(5)=5, g(6)=4, f′(4)=6, f′(5)=5, f′(6)=4, g′(4)=5, g′(5)=5, g′(6)=4 a) 16 b) 19 c) 18 d) 17 e) 15

  4. clara1223
    • one year ago
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    sorry, thanks for all your help, im just on a time crunch.

  5. Jhannybean
    • one year ago
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    You're not taking a test are you?

  6. clara1223
    • one year ago
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    no, I just have to be somewhere soon

  7. Jhannybean
    • one year ago
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    Oh, alright.

  8. clara1223
    • one year ago
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    And this homework is due tomorrow

  9. Jhannybean
    • one year ago
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    So i assume : \[(g~\circ ~ f)'(x) \qquad \implies \qquad \frac{d}{dx}(g(f(x)) = g'(f(x)) \cdot f'(x)\]

  10. Jhannybean
    • one year ago
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    where \(x=6\)

  11. Jhannybean
    • one year ago
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    Ack, I've got to head off for a little while, I will be back very soon.

  12. Jhannybean
    • one year ago
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    @DanJS @jim_thompson5910 mind taking over? Thanks!

  13. DanJS
    • one year ago
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    i am late, are you still working on these things?

  14. DanJS
    • one year ago
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    yes, that one would be just the chain rule, and they list the values you may need for the functions and their derivatives \[\qquad \frac{d}{dx}(g(f(6)) = g'(f(6)) \cdot f'(6) \]

  15. DanJS
    • one year ago
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    f(6) = 6 f '(6) = 4 g'(6)=4 so \[\qquad \frac{d}{dx}(g(f(x)) = g'(f(x)) \cdot f'(x) = g'(6)*f'(6) = 4*4\]

  16. DanJS
    • one year ago
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    To find f'(x) you use the chain rule... \[\frac{ d }{ dx }f(x) = \frac{ d }{ du }*\frac{ du }{ dx } f(x)\]

  17. DanJS
    • one year ago
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    If you have a function inside a function , like f(g(x)) f(x) = sin(x^3) I usually think something like; " f ' (x) is equal to derivative of sin(inside function left alone) times the derivative of the inside function" f ' (x) = cos(x^3) * 2x^2 thats it, goodluck

  18. DanJS
    • one year ago
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    similar with the given function f(x) = 8*sin(x^(1/2))^3 start with outside and leave inside function same, except here you have to use the rule two times, u^3, sin(u) and x^(1/2) start from outside power function \[f ' (x) = 3*8*[\sin(x^{1/2})]^2~*~\cos(x^{1/2})*\frac{ 1 }{ 2 }*x^{-1/2}\]

  19. DanJS
    • one year ago
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    simplifies to D i think ... check it out,

  20. DanJS
    • one year ago
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    keep leaving the inner functions alone, and work outside to inside, power rule - derivative of sin() - derivative of root x

  21. DanJS
    • one year ago
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    \[[f(g(h(x)))] ' = f'(g(h(x))) *g'(h(x)) * h'(x)\]

  22. DanJS
    • one year ago
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    you can have as many nested functions as you want, same idea

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