Question

How do you find f'(x) using the definition of a derivative for `f(x)=x^3 + 2x^2 + 1`?

Answer 1

The derivative is `3 x^2 + 4x`

Explanation:

The definition is:

`f'(x) = lim_{h \to 0} {f(x+h) - f(x)}/h`

So the terms are:

`f(x+h)=(x+h)^3 + 2(x+h)^2 + 1`.

Term by term, we have:

1. `(x+h)^3=x^3+3 x^2 h+3 x h^2+h^3`
2. `2(x+h)^2 = 2(x^2+2hx+h^2)=2x^2+4hx+2h^2`

Obviously,

`f(x)=x^3+2x^2+1`.

Now we compute `f(x+h)-f(x)`, since it's very long, I'll just write down the result. If you'll need help with calculations don't hesitate to ask (but I strongly suggest to do them yourself, since it will improve your skill):

`2 h^2+h^3+4 h x+3 h^2 x+3 h x^2`

Now we divide this quantity by `h`:

`2 h+h^2+4 x+3 h x+3 x^2`.

Now, taking the limit of this quantity as `h\to 0` means to erase all the terms involving `h`, leaving `3 x^2 + 4x` as the final result.