Find f’(2) for f(x)= 1/x-3 using the definition of the derivative

Answer 1

Given: f(x) = 1/(x-3) = (x-3)^-1

The definition of derivative in terms of limits is:

f'(x) = lim _h→0 (f(x+h) - f(x))/h

f(x-h) = 1/(x-h-3)

∴f'(x) = lim _h→0 (1/(x+h-3) - 1/(x-3))/h

LCD is (x+h-3)(x-3)

f'(x) = lim _h→0 ((x-3-(x+h-3)/(x+h-3)(x-3))/h

Distribute the negative:

f'(x) = lim _h→0 ((x-3-x-h+3)/(x+h-3)(x-3))/h

Combine like terms:

f'(x) = lim _h→0 (-h/(x+h-3)(x-3))/h

Use division of fractions (keep, change, flip)

f'(x) = lim _h→0 -1/(x+h-3)(x-3)

Now you can plugin 0 for h

f'(x) = -1/(x+0-3)(x-3)

f'(x) = -1/(x-3)²

To check, do the chain rule:

f'(x) = -(x-3)^-2 = -1/(x-3)²

Answer 2

This is a chain rule problem that is not totally obvious. The outer function is h(x) = 1/x and the inner function is g(x) = x-3.

This is because h(g(x)) = 1/(x-3).

It is easier to use power rule for the outer function when you use negative exponents. Re-write the function:

f(x) = 1/(x-3) = (x-3)^-1

So, the derivative of f(g(x)) = f '(g(x))*g'(x).

f '(x) = -1(x-3)^-2*d/dx(x-1)

f ' (x) = -1/[(x-3)^2] *1

so, plug 2 into f '(x)!

Hope this helps!