1. ## Differentiation

This is a topic i am really struggling to get my head around.

1. the expression (x+a)^-1 can be approximated for small values of a by the relationship

(x+a)^-1 = (1/x) - (a/x^2) + (a^2/x^3)

Using the relationship, derive an expression for the rate of change of the function f(x)=1/x by the 'first principles' differentiation method.

2. Originally Posted by daniel123
This is a topic i am really struggling to get my head around.

1. the expression (x+a)^-1 can be approximated for small values of a by the relationship

(x+a)^-1 = (1/x) - (a/x^2) + (a^2/x^3)

Using the relationship, derive an expression for the rate of change of the function f(x)=1/x by the 'first principles' differentiation method.
f'(x)=lim $\frac{f(x+a)-f(x)}{(x+a)-x}$, where a-> 0

now substitute f(x+a) by $(x+a)^{-1} = \frac{1}{x} - \frac{a}{x^2} + \frac{a^2}{x^3}$ and f(x) by 1/x in the f'(x) formula.

3. Ok thankyou, i have an answer, which is,

=-1/x^2