2007-08-20 19:17:50 · 5 answers · asked by M N 1 in Science & Mathematics ➔ Mathematics
in getting the derivative of the function
y = (2x^3-5x)^(1/4)
applying the chain rule
it will become
y' = 1/4 (2x^3 - 5x)^(1/4-1) d (2x^3-5x)
y' = 1/4 (2x^3 - 5x)^ -3/4 (6x^2 - 5)
y' = 3/2 x^2 - 5/4 (2x^3 - 5x)^ -3/4
hope it helps... =)
2007-08-20 19:25:25 · answer #1 · answered by >bLueeyes< 2 · 0⤊ 0⤋
M.Jith does it 90% correct... but mess it up during Chain-Rule
the answer is : (1/4(2x^3-5x)^(-3/4))*(6x^2-5)
2007-08-20 19:26:56 · answer #2 · answered by John N 2 · 0⤊ 0⤋
f (x) = (2x³ - 5x)^(1/4)
f `(x) = (1/4) (2x³ - 5x)^( - 3/4) (6x² - 5)
f `(x) = (6x² - 5) / [4 (2x³ - 5x)^(3/4) ]
2007-08-20 19:32:21 · answer #3 · answered by Como 7 · 3⤊ 0⤋
y' = 1/4(2x^3 - 5x)^(-3/4) ( 2x^3-5x)'
y'= 1/4(2x^3 - 5x) ^(-3/4) ( 6x^2 - 5)
2007-08-20 23:06:34 · answer #4 · answered by Trần Hoài Vũ 2 · 0⤊ 2⤋
dy/dx = 1/4 (2x^3 - 5x)^(-3/4) * (6x^2 - 5)
power rule + chain rule --> d(u^n)/dx = n(u^(n-1)) * du
yeah thanks john.. u r right.
2007-08-20 19:22:25 · answer #5 · answered by Anonymous · 0⤊ 1⤋