Question

Using the limit definition, how do you find the derivative of `f(x)= 2x^2-x`?

Answer 1

`f'(x)=4x-1`

Explanation:

By limit definition, the derivative of any function is
`f'(x)=lim_(h\to0) (f(x+h)-f(x))/h`

Now, for this particular question, we have `f(x)=2x^2-x`
So, using the derivative definition, we have
`f'(x)=lim_(h\to0)((2(x+h)^2-(x+h))-(2x^2-x))/h`

Rearranging the equation and separating to have the `x` parameters of same power, we have
`f'(x)=lim_(h\to0)((2(x+h)^2-2x^2)/h-(x+h-x)/h)`

In the right side, the second term is simple so we'll directly get it as `1`. In the first term, notice that we can take `2` as a common term. Next, by taking `a=x+h` and `b=x`, we can use the identity `a^2-b^2=(a+b)(a-b)` for which we get the answer `(2x+h)(h)` Substituting that above, we get
`f'(x)=lim_(h\to0)(2(2x+h)h/h-1)=lim_(h\to0)(2(2x+h)-1)`

Now, since it's not possible to reduce the terms any further, we can finally just take `h=0` and get what we wanted. Hence you should get the answer I've written way above.