I can know how to differentiate it and i also know how to use the limit definition of derivative to use to solve x^n where n is positive but not when n is negative. Thanks for your help!
-
The limit definition: f'(x) = Lim ∆x --> 0 [f(x + ∆x) - f(x)]/∆x
f(x + ∆x) = (x + ∆x)^-2 = 1/(x + ∆x)²
f(x) = x^-2 = 1/x²
f'(x) = lim ∆x --> 0 [1/(x + ∆x)² - 1/x²]/∆x
multiply through by x²*(x + ∆x)² to get
f'(x) = lim ∆x --> 0 [x² - (x + ∆x)²] / ∆x*x²*(x + ∆x)²
f'(x) = lim ∆x --> 0 [-2*x*∆x - ∆x²] / [∆x*x^4 + 2*x³*∆x²+ ∆x³]
f'(x) = lim ∆x --> 0 [-2*x - ∆x] / [x^4 + 2*x³*∆x+ ∆x²]
let ∆x = 0
f'(x) = -2*x / x^4 = -2/x³ = -2*x^-3
f(x + ∆x) = (x + ∆x)^-2 = 1/(x + ∆x)²
f(x) = x^-2 = 1/x²
f'(x) = lim ∆x --> 0 [1/(x + ∆x)² - 1/x²]/∆x
multiply through by x²*(x + ∆x)² to get
f'(x) = lim ∆x --> 0 [x² - (x + ∆x)²] / ∆x*x²*(x + ∆x)²
f'(x) = lim ∆x --> 0 [-2*x*∆x - ∆x²] / [∆x*x^4 + 2*x³*∆x²+ ∆x³]
f'(x) = lim ∆x --> 0 [-2*x - ∆x] / [x^4 + 2*x³*∆x+ ∆x²]
let ∆x = 0
f'(x) = -2*x / x^4 = -2/x³ = -2*x^-3