Cannot seem to figure out how to do this type of problem. Any help would be appreciated.
Y=(2 + 2/X) ^1/8
2007-10-06 01:14:10 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Y=(2 + 2/X) ^1/8
Y'=1/8(2+2/x)^(-7/8) *(-2/X^2)
Y'=[(2+2/x)^(-7/8) ]/(-1/4X^2)
2007-10-06 01:19:55 · answer #1 · answered by ptolemy862000 4 · 1⤊ 0⤋
Calculus right? Hey... Don't worry, it gets pretty hard sometimes but just keep on trying and never give up. Math is easy once you get the hang of it.
Here's my answer but I'm not 100% sure ok? I used rule number eight. The GENERAL POWER RULE.
y=(2+2/x)^1/8
y'=1/8(2+2/x)^-7/8 * (2)
y'=1/4(2+2/x)^-7/8
There you have it. If you want, you can bring down (2+2/x)^-7/8 to make the exponent positive. Also, you can use a square root on (2+2/x) ... :-) I'm not sure okay? I'm just 15 and just in 4th year high school... So... Just try and check it.
2007-10-06 01:35:33 · answer #2 · answered by Rinko_Sano 2 · 0⤊ 0⤋
y' = 1/8 (2 +2/x)^-â ( -2/ x²)
2007-10-06 02:29:01 · answer #3 · answered by mramahmedmram 3 · 0⤊ 0⤋
y = (2 + 2/x)^(1/8)
dy/dx = (1/8)(2 + 2/x)^( - 7/8 ) (- 2 / x ²)
dy/dx = (- 1 ) (2 + 2/x)^( - 7/8 ) / (4 x ²)
2007-10-06 06:06:23 · answer #4 · answered by Como 7 · 1⤊ 0⤋