Abmon98
  • Abmon98
Find the nth derivative of 4/(6x+8)^3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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AihberKhan
  • AihberKhan
http://www.wolframalpha.com/input/?i=Find+the+nth+derivative+of+4%2F%286x%2B8%29%5E3 That might help.
Abmon98
  • Abmon98
i find a problem with writing factorials in the formula. f'(x)=4(-3)*6(6x+8)^-4 f''(x)=4(-3)(-4)(6)^2(6x+8)^-4, i recognize the pattern but cant interpret it into a formula
mathmate
  • mathmate
@Abmon98 It is much easier to differentiate the function two or three times and find the relationship from what you got from the Wolfram link. This will help you see how things actually work. Remember, you need to use the quotient rule, and the chain rule. As a hint, I will provide you for checking: f(x)=4/(6x+8) f'(x)=-24/(6x+8)^2 f"(x)=288/(6x+8)^3 .... Actually take the time to work them out and figure out and deduce the relationship.

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Abmon98
  • Abmon98
the function is given y=4/(6x+8)^3 y'=4(-3)(6)/(6x+8)^4 y''=4(-3)(-4)(6)^2/(6x+8)^5 y'''=4(-3)(-4)(-5)6^3/(6x+8)^6
Abmon98
  • Abmon98
in the nth derivative formula 6 is raised to the nth power and the constant 4 remains as it is. I am struggling with factorials.
Zarkon
  • Zarkon
\[\frac{(-1)^n(n+2)!}{2}\]
Abmon98
  • Abmon98
why is it n+2!
Zarkon
  • Zarkon
for n=1 you get -3, 1+2=3 for n=2 you get (-3)(-4), 2+2=4
Abmon98
  • Abmon98
what about (-1)^n+2(n+2)!
Zarkon
  • Zarkon
well that doesn't work...so no
Abmon98
  • Abmon98
so is the formula of the nth derivative is (-1)^n(n+2)!6^n*4/(6x+8)^n+3
mathmale
  • mathmale
Why not check out the formula above by actually using it to find the 1st, 2nd and 3rd derivatives? Does the formula produce the results you expected?
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle y(x)=\frac{a}{(bx+c)^k} }\) (of course k>1) \(\color{#000000 }{ \displaystyle y(x)=a(bx+c)^{-k} }\) \(\color{#000000 }{ \displaystyle y'(x)=(-k)ab(bx+c)^{-k-1} }\) \(\color{#000000 }{ \displaystyle y'(x)=k(-1)^1ab(bx+c)^{-k-1} }\) \(\color{#000000 }{ \displaystyle y''(x)=k(-1)^1(-k-1)ab^2(bx+c)^{-k-2} }\) \(\color{#000000 }{ \displaystyle y''(x)=k(-1)^1(-1)^1(k+1)ab^2(bx+c)^{-k-2} }\) \(\color{#000000 }{ \displaystyle y''(x)=k(k+1)(-1)^2ab^2(bx+c)^{-k-2} }\) \(\color{#000000 }{ \displaystyle y'''(x)=k(k+1)(-1)^2(-k-2)ab^3(bx+c)^{-k-3} }\) \(\color{#000000 }{ \displaystyle y'''(x)=k(k+1)(k+2)(-1)^3ab^3(bx+c)^{-k-3} }\) \(\color{#000000 }{ \displaystyle y^{(4)}(x)=k(k+1)(k+2)(-1)^3(-k-3)ab^4(bx+c)^{-k-4} }\) \(\color{#000000 }{ \displaystyle y^{(4)}(x)=k(k+1)(k+2)(k+3)(-1)^4ab^4(bx+c)^{-k-4} }\) So you can tell that, \(\color{#000000 }{ \displaystyle y^{(\color{blue}{n})}(x)=\frac{(k+\color{blue}{n})!}{k!} (-1)^\color{blue}{n}ab^\color{blue}{n}(bx+c)^{-k-\color{blue}{n}} }\) and let me correct this for non-integer k, \(\color{#000000 }{ \displaystyle y^{(\color{blue}{n})}(x)=\frac{\Gamma (k+1+\color{blue}{n})}{\Gamma (k+1)} (-1)^\color{blue}{n}ab^\color{blue}{n}(bx+c)^{-k-\color{blue}{n}} }\)
SolomonZelman
  • SolomonZelman
Let me check this, hold on please
SolomonZelman
  • SolomonZelman
Yes, there is an error in last two equations...
SolomonZelman
  • SolomonZelman
It actually comes down to, \(\color{#000000 }{ \displaystyle y^{(\color{blue}{n})}(x)=\frac{(k-1+\color{blue}{n})!}{(k-1)!} (-1)^\color{blue}{n}ab^\color{blue}{n}(bx+c)^{-k-\color{blue}{n}} }\) and then I will correct this for non-integer k, the following way, \(\color{#000000 }{ \displaystyle y^{(\color{blue}{n})}(x)=\frac{\Gamma (k+\color{blue}{n})}{\Gamma (k)} (-1)^\color{blue}{n}ab^\color{blue}{n}(bx+c)^{-k-\color{blue}{n}} }\)

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