a) 1/(2x^2) + 1/(3x^3)
b) (x-1)/x^(1/2)
2006-12-31 04:41:58 · 7 answers · asked by Anonymus 2 in Science & Mathematics ➔ Mathematics
a) 1/(2x^2) + 1/(3x^3) = y
y = (1/2)x^-2 + (1/3)x^-3
y' = -1/(x^3) - 1/(x^4)
b) (x-1)/x^(1/2) = y
y = √x - 1/√x
y' = 1/(2√x) + 1/[2x^(3/2)]
safe
2006-12-31 22:02:31 · answer #1 · answered by matt.colemanx 1 · 0⤊ 0⤋
a) 1/(2x^2) + 1/(3x^3) = y
y = (1/2)x^-2 + (1/3)x^-3
y' = -1/(x^3) - 1/(x^4)
b) (x-1)/x^(1/2) = y
y = √x - 1/√x
y' = 1/(2√x) + 1/[2x^(3/2)]
2006-12-31 04:51:53 · answer #2 · answered by sahsjing 7 · 1⤊ 0⤋
y=1/2. x^-2+ 1/3. x^-3
y'=1/2.(-2).x^-3+ 1/3.(-3)x^-4
y'=-x^-3-x^-4
=1/x^3-1/x^4
2006-12-31 04:48:44 · answer #3 · answered by iyiogrenci 6 · 0⤊ 0⤋
(People normally call 'log to the base of e' as 'ln'.) To differentiate xln(x), we use the product rule: (Let u = x; ln(x) = v.) Let f(x) = xln(x) f'(x) = u*v' + v'*u = x(1/x) + ln(x)(x) =1 + xln(x) There is a formula for differentiating terms with ln. For y = ln(u), y' = u / u'
2016-05-22 23:48:45 · answer #4 · answered by Anonymous · 0⤊ 0⤋
a) -1/x^3 - 1/x^4
b)[1/2x^(1/2)](1+1/x)
2006-12-31 04:46:47 · answer #5 · answered by Mena M 3 · 0⤊ 1⤋
I can't diffentiate but all respect to X anyway!
2006-12-31 04:47:36 · answer #6 · answered by Kate J 4 · 0⤊ 2⤋
f(x)=(1/2)x^-2+(1/3)x^-3
f'(x)=(1/2)(-2)x^-3+(1/3)(-3)x^-4
=-1/x^3-1/x^4
b.f(x)=x^1/2-x^-1/2
f'(x)=1/2x^-1/2-(-1/2)x^-3/2
=1/2x^1/2+1/2x^3/2
2006-12-31 05:45:38 · answer #7 · answered by raj 7 · 0⤊ 0⤋