Question

How do you find the derivative using limits of `f(x)=x^3+x^2`?

Answer 1

`f'(x) = 3x^2 +2x`

Explanation:

The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So with `f(x) = x^3+x^2` then;

`f(x+h) = (x+h)^3+(x+h)^2`
`\ \ \ \ \ \ \ = (x^3+3x^2h+3xh^2+h^3) +(x^2+2xh+h^2)`

And so:

`f(x+h)-f(x) = x^3+3x^2h+3xh^2+h^3 +x^2+2xh+h^2 -x^3-x^2`
`\ \ \ \ \ \ \ = 3x^2h+3xh^2+h^3 +2xh+h^2`

And so the derivative of `f(x)` is given by:

`f'(x) = lim_(h rarr 0) ( f(x+h)-f(x) ) / h`
`\ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 3x^2h+3xh^2+h^3 +2xh+h^2 ) / h`
`\ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 3x^2+3xh+h^2 +2x+h )`
`\ \ \ \ \ \ \ \ = 3x^2 +2x`