Finding the derivative?

Question

f(x)=1/x^6-1/x^8 using negative powers of x.
I know the answer is f '(x)=-6/x^2+8/x^9 but how do I get it?

Answer 1

1) Given f(x) = 1/x^6 - 1/x^8

Rewrite the given expression in negative powers of x

f(x) = x^-6 - x^-8

Differentiating with respect to x

f'(x) = d/dx(x^-6 - x^-8)

f'(x) = d/dx(x^-6) - d/dx(x^-8)

Recall the formula of differentiation : d/dx(x^n) = n*x^n-1

f'(x) = -6*x^-6-1 - ( - 8*x^-8-1)

f'(x) = -6*x^-7 - ( -8*x-9 )

f'(x) = -6*x^-7 + 8*x^-9

Rewriting the expression inpositive powers of x

f'(x) = -6/x^7 + 8/x^9

Answer 2

Given f(x) = 1 / x⁶ - 1/ x⁸

Rewrite the given expression in negative powers of x

f(x) = x^-6- x^-8

Diferenciate with respective x

f ' (x) = d / dx(x^-6) - d / dx(x^-8)

Recall : Differciation formula d / dx (xⁿ) = nx^n-1

f ' (x) = (-6x^-6-1) - ( -8x^-8-1)

         = (-6x^-7) - (-8x^-9)

         = (-6 / x⁷) - (-8 / x⁹)

         = (-6 / x⁷) + (8 / x⁹)

Given answer is wrong

f ' (x) = (-6 / x⁷) + (8 / x⁹).