the derivative of (6/5)x^(5/6)
is
x^4?
2007-05-26 16:45:35 · 10 answers · asked by john d 2 in Science & Mathematics ➔ Mathematics
This may be easier to understand if you use the product rule since there are actually two numbers here. The product rule says this:
Given u = the product of v and w, then u' = v w' + w v'.
If we let u = (6/5)x^(5/6), then v = 6/5, and w = x^(5/6), and u'
= (6/5)[(5/6)x^(5/6 - 6/6)] + [x^(5/6)](0). Simplifying this further, we get:
u' = (1)x^(- 1/6) + 0
u' = x^(-1/6).
So x^(-1/6) is the derivative of your function.
2007-05-26 17:14:24 · answer #1 · answered by MathBioMajor 7 · 0⤊ 0⤋
No, using the power rule the derivative is (6/5)(5/6)x^(5/6 - 1)
or x^(-1/6)
2007-05-26 23:48:43 · answer #2 · answered by Math Nerd 3 · 0⤊ 0⤋
nope
derivative of x^n = n x^(n-1)
right?
in your case n = 5/6 so n-1 = -1/6
answer is
x^(-1/6)
or...
1/x^(1/6)
2007-05-26 23:49:12 · answer #3 · answered by electric 3 · 0⤊ 0⤋
No. It is x^(-1/6).
2007-05-26 23:47:54 · answer #4 · answered by Zhuo Zi 3 · 1⤊ 0⤋
No, answer is x^-(1/6)
2007-05-26 23:54:55 · answer #5 · answered by dongskie mcmelenccx 3 · 0⤊ 0⤋
(d/dx)((6/5)x^(5/6))=(6/5)*(5/6)x^((5/6)-1)
(since derivative of x^n is (n*x^(n-1))
Therefore L.H.S =1*x^((5-6)/6)
L.H.S = x^(-1/6)
L.H.S = (1/x^6)
Therefore your answer is not correct.......
2007-05-26 23:57:37 · answer #6 · answered by sriram t 3 · 0⤊ 0⤋
Nope.
(ask NASA)...lol
2007-05-26 23:53:19 · answer #7 · answered by Sanjiv 3 · 0⤊ 0⤋
no
2007-05-26 23:55:07 · answer #8 · answered by Shane 4 · 0⤊ 0⤋
no
2007-05-26 23:47:39 · answer #9 · answered by Jack Flash 4 · 0⤊ 0⤋
oh man......is that algebra?
2007-05-26 23:51:29 · answer #10 · answered by frank l 1 · 0⤊ 0⤋