# Getting from difference quotient, to the polynomial "shortcut"

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.

• "Linearity of differentiation. In calculus, the derivative of any linear combination of functions equals the same linear combination of the derivatives of the functions" May 20, 2017 at 4:08
• 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 $g$. May 20, 2017 at 5:02

${\frac {{\mbox{d}}}{{\mbox{d}}x}}(\alpha \cdot f(x)+\beta \cdot g(x)) = {\frac {{\mbox{d}}}{{\mbox{d}}x}}(\alpha \cdot f(x))+{\frac {{\mbox{d}}}{{\mbox{d}}x}}(\beta \cdot g(x))$
A good practice problem to illustrate this point would be to take the derivative with respect to x of any polynomial: $a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{2}x^{2}+a_{1}x+a_{0}$