Differentiate: (x+1/x)[(x)sq.root+1/(x)sq.root]power3, x not =0?

Answer 1

differentiating u v, we get u' v + u v'
apply the same

diff [x+1/x]

you get [1 - 1/x²]

diff [√x + 1/√x]³

you get 3[x + 1/x][√x + 1/√x]² [1/(2√x) - 1/{2(√x)³}]

so the answer is
[1 - 1/x²][√x + 1/√x]³ + 3[x+1/x][x + 1/x][√x + 1/√x]² [1/(2√x) - 1/{2(√x)³}]

Answer 2

f(x) = a million/x f(x+h) = a million/(x+h) f(x+h) - f(x) = a million/(x+h) - a million/x multiply RHS with x(x+h) / x(x+h) f(x+h) - f(x) = (x- (x+h))/ x(x+h) => -h/x(x+h) dividing by ability of h [ f(x+h) - f(x) ]/ h = -h / x(x+h)h => -a million/x(x+h) => -a million/(x^2 + xh) while h -> 0 , [a million/(x^2 + xh)] -> -a million/x^2 f'(x) = lim(h->0) [f(x+h)-f(x)]/h = (-a million/x^2)

Answer 3

y'=(1-1/x^2)((sqrt(x)+1/sqrt(x))

+( (1/sqrt(x)-(1/2)(x^(-3/2) ) (x+1/x) )

Answer 4

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