Quote:
Originally Posted by choiminho
using the first principle of differentiation, find the first derivatives of
1.f(x) = 3/x^2
2. 1/(sqrt x)^3
-i don't know how to using limit to solve this. Please help me.  |
1. Using the quotient rule to solve:
f'(x) = ((derivative of the top)(bottom) - (top)(derivative of the bottom))/(bottom squared)
f'(x) = ((0)(x^2) - (3)(2x))/(x^4) = -6x/x^4 = -6/x^3
Using the product rule to solve:
3/x^2 = 3x^-2
f'(x) = (derivative of first)(second) + (first)(derivative of second)
f'(x) = (0)(x^-2) + (3)(-2x^-3) = -6x^-3 = -6/x^3
Using limit method:
limit as h goes to 0 of (3/(x+h)^2 - 3/x^2)/h = ((3x^2 - 3(x + h)^2)/((x^2)(x + h)^2))/h
= (3x^2 - 3x^2 - 6xh - 3h^2)/((h)(x + h)^2(x^2))
= (-3h(2x + h))/((h)(x + h)^2(x^2))
= (-3(2x + h))/((x + h)^2(x^2))
Now take the limit as h goes to 0.
= (-6x)/(x^4) = -6/x^3
2. f(x) = 1/(sqrt x)^3 = (sqrt(x))^-3 = (x^(1/2))^-3 = x^(-3/2)
f'(x) = (-3/2)(x^(-3/2 - 1) = (-3/2)(x^(-5/2) = -3/(2(sqrt(x))^5)