Now find the derivative of the following function using first principles correctly

Question

Now find the derivative of the following function using first principles correctly

f(x)=x^3-3x

Answer 1

The way I interpret the question, I believe that finding the derivative using first principles means that the question wants us to use the limit definition of the derivative. There are two different forms, but I would suggest using the following form if you're allowed to choose which one to use:

f'(a)=lim_x->a (f(x)-f(a))/(x-a)

Then we can plug in the function given in the problem:

f'(a)=lim_x->a((x^3-3x)-(a^3-3a))/(x-a)

Simplifying the numerator gives us:

f'(a)=lim_x->a(x^3-3x-a^3+3a)/(x-a)

We want to eventually cancel out the term (x-a) from the numerator, so now we can factor out the numerator.

Looking just at the numerator, we have

x^3-3x-a^3+3a

Regrouping the terms, we can get

(x^3-a^3)+(-3x+3a)

You might notice that the first two terms and the last two terms both make expressions that are divisible by x-a. If not, I will explain why this is true.

For the first two terms, we can use the identity (a^3-b^3)=(a-b)(a^2+ab+b^2). For the last two terms, we can divide both terms by -3 to get -3x+3a=-3(x-a).

This means that we can rewrite the numerator as

(x-a)(x^2+ax+a^2)+(-3)(x-a)

Using the expression we had above, this means we can say

f'(a)=lim_x->a(x^3-3x-a^3+3a)/(x-a)= lim_x->a((x-a)(x^2+ax+a^2)-3(x-a))/(x-a)

Cancelling out the x-a term gives us

f'(a)=lim_x->a(x^2+ax+a^2-3)

Now we can find the limit by plugging in a wherever there is an x:

f'(a)=a^2+a(a)+a^2-3=3a^2-3

Since we want f'(x), we can just substitute x for a, so the final answer is

f'(x)=3x^2-3

(If you've learned some of the basic differentiation rules, you can check that you would get this answer by using the power rule and the constant multiple rule)

I know it can be hard to read with everything typed out like that, so please let me know if you need any clarification on what I've written! Hope this helps!