Question

Find the nth derivative of 4/(6x+8)^3

Answer 1

http://www.wolframalpha.com/input/?i=Find+the+nth+derivative+of+4%2F%286x%2B8%29%5E3

That might help.

Answer 2

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

Answer 3

@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.

Answer 4

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

Answer 5

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.

Answer 6

\[\frac{(-1)^n(n+2)!}{2}\]

Answer 7

why is it n+2!

Answer 8

for n=1 you get -3, 1+2=3
for n=2 you get (-3)(-4), 2+2=4

Answer 9

what about (-1)^n+2(n+2)!

Answer 10

well that doesn't work...so no

Answer 11

so is the formula of the nth derivative is (-1)^n(n+2)!6^n*4/(6x+8)^n+3

Answer 12

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?

Answer 13

\(\color{#000000 }{ \displaystyle y(x)=\frac{a}{(bx+c)^k} }\) (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,

\(\color{#000000 }{ \displaystyle y^{(\color{blue}{n})}(x)=\frac{\Gamma (k+1+\color{blue}{n})}{\Gamma (k+1)} (-1)^\color{blue}{n}ab^\color{blue}{n}(bx+c)^{-k-\color{blue}{n}} }\)

Answer 14

Let me check this, hold on please

Answer 15

Yes, there is an error in last two equations...

Answer 16

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,

\(\color{#000000 }{ \displaystyle y^{(\color{blue}{n})}(x)=\frac{\Gamma (k+\color{blue}{n})}{\Gamma (k)} (-1)^\color{blue}{n}ab^\color{blue}{n}(bx+c)^{-k-\color{blue}{n}} }\)