3

I wanted to resolve this problem : $$ f(x) = 3 x^2 - 5 x - 2 $$

to a derivative, and I did it like this :

My work

$ \begin{align} f(x) &= x^3-5x-2 \\ f'(x) &= 3x^2-5-2 \\ &= 3x^2-7 \end{align} $

but once I checked the correction, I found this: $$ f'(x) = 3 x^2 - 5 $$

And I really don't know how they manage to get $3x^2 -5$ instead of $3x^2-7$.

Thanks for you time

R00t
  • 35
  • 1
    To see why the derivative of a constant function is zero, you should think about the interpretation of the derivative. – Nikolaj-K Jan 21 '15 at 18:16

3 Answers3

7

The derivative of a constant is zero, so if $$f(x) = x^3 - 5x - 2 = x^3 - 5 x^1 - 2,$$ the correct derivative is $$f'(x) = 3x^{3-1} - 5 \cdot 1 x^{1-1} - \color{red}{0} = 3x^2 - 5x^0 - 0 = 3x^2 - 5.$$

heropup
  • 135,869
4

To take a derivative, recall the the operation on a polynomial is the following: \begin{align} \frac{d}{dx}\left(x^n\right)=nx^{n-1},\tag{1} \end{align} meaning that in your problem, the derivative is this: \begin{align} \frac{d}{dx}\left(x^3-5x-2\right)=3x^2-5.\tag{2} \end{align} Also, realize that a constant, such as "$-5$" has no rate of change, and therefore its derivative is $0$.

bjd2385
  • 3,017
2

The derivative of a constant is zero. It was your mistake.

Alex Silva
  • 3,557