In this MIT introduction lecture to derivatives, the professor takes us through how the difference quotient can produce a formula for quickly finding derivatives for $x^n$ where $n = 1, 2, 3... $

He then jumps to saying it works for polynomials. I don't see how this: (1)

$$\frac{d}{dx}x^n = nx^{n-1} $$

Can justifiably be used on polynomials without further explanation (although I see, and can re-apply what he did to the polynomial).

I had a go at solving this, but only managed to repeat his process for $x^n $ to $ax^n$. But I did not know where to go from there to have it apply to any polynomial.

Have I just not spotted how (1) can be justifiably used for polynomials? Or does the way in which he find the differential of the polynomial rely on a (far) more complicated extension of the difference quotient?

The reason I ask, (since what I've said might not make sense) is I just don't understand the jump from (1) to using it across all terms of a polynomial.

  • $\begingroup$ "Linearity of differentiation. In calculus, the derivative of any linear combination of functions equals the same linear combination of the derivatives of the functions" $\endgroup$
    – Biggs
    May 20, 2017 at 4:08
  • 1
    $\begingroup$ In particular, you can work out from the difference quotient that the derivative of $f+g$ equals the derivative of $f$ plus the derivative of $g$. $\endgroup$ May 20, 2017 at 5:02

1 Answer 1


Linearity of differentiation means (to put it plainly) that the operation of taking a derivative can be applied to each term within the polynomial separately.

${\frac {{\mbox{d}}}{{\mbox{d}}x}}(\alpha \cdot f(x)+\beta \cdot g(x)) = {\frac {{\mbox{d}}}{{\mbox{d}}x}}(\alpha \cdot f(x))+{\frac {{\mbox{d}}}{{\mbox{d}}x}}(\beta \cdot g(x))$

This principle is enough to apply the derivate rules to a general polynomial.

A good practice problem to illustrate this point would be to take the derivative with respect to x of any polynomial: $a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{2}x^{2}+a_{1}x+a_{0}$

Hope this helps!

  • $\begingroup$ Thank you, +1 for the extra help. (Although I still need to digest it all, it's several helpful starting points for further reading if I need to). But one brief clarification if you don't mind: does your main equation demonstrate the linearity of differentiation? Or does the equation follow from knowing the linearity of differentiation? $\endgroup$
    – Jodes
    May 20, 2017 at 16:16
  • $\begingroup$ The main equation above demonstrates the linearity of differentiation, but it also follows for a polynomial by "knowing" the linearity of differentiation. The reason they are the same is that generally, the Linearity of Differentiation means "the derivative of any linear combination of functions equals the same linear combination of the derivatives of the functions". An n degree polynomial can be described as a linear combination of the functions 1, x, x^2, ... x^n, etc. Reflect on this and you can see why the main equation is both a theorem and an example, if you will :) $\endgroup$ May 20, 2017 at 20:09

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