Finding the slope of a tangent to the graph of a function is one of the most basic applications of the notion of derivative. For some curves, it is possible to find the derivative without resorting either to computing the limits or employing previously proven rules for computing derivative.

## Quadratics

A tangent line to a parabola f(x) = ax^2 + bx + c has the property of meeting the parabola at a single point. This means that, for a particular m , the system

has a unique solution. Since the system reduces to a quadratic equation, it has a unique solutions provided its discriminant vanishes. So solve

for x:

Since the solution must be unique, b^2 - 4a(c - B) = 0, implying

and, therefore, m = 2ax + b. Thus f '(x) = 2ax + b, as expected.

## Reciprocals

Now consider the reciprocal function f(x) = a/x. As before, we arrive at the equation a/x = mx + B, which leads to the quadratic equation

with the solution

Assuming the discriminant 0, x = -B / 2m. Thus, B = -2mx, and since

we have

## Square Root

This time, f(x) = \sqrt{x}. Again, \sqrt{x} = mx + B.

Squaring both sides leads to a quadratic equation: m^2x^2 + (2mB - 1)x + B^2 = 0, with a solution

Simplifying we get

Since the discriminant must be zero, B = 1/4m, and therefore x = (1 - 2mB) / 2m^2 = 1 / 4m^2, and since x \gt 0, m = 1/2\sqrt{x}. Hence, f '(x) = 1 / 2\sqrt{x}.

## Any Polynomial

Write p(x+h) as a polynomial in h with coefficients that depend on x. The coefficient of the linear (in h) term is the derivative p'(x). For example, (x+h)^2=x^2+2xh+h^2, so ({x^2})'=2x. By the binomial formula (x+h)^n = x^n + nx^{n-1}h +(\dots)h^2 , whence (x^n)'=nx^{n-1} .

Another way to calculate p'(x) is to divide p(x)-p(a) by x-a, which is possible, because p(a)-p(a)=0, and stick x=a into the answer. For example, x^3-a^3=(x^2+xa+a^2)(x-a), so ({x^3})'=3x^2.

## Root of Degree n.

To calculate the derivative of \sqrt[n]{x} we have to make sense of the difference quotient (\sqrt[n]{x} - \sqrt[n]{a})/(x-a) for x=a. This can be written as (y-b)/(y^n-b^n), where y=\sqrt[n]{x} and b=\sqrt[n]{a}. It is the same problem as for the monomial y^n, but "upside down," so we factor the denominator instead of the numerator, and get 1/(y^{n-1} + y^{n-2}b + \dots + y b^{n-2}+b^{n-1}) , that makes sense for y=b and gives us the answer: (\sqrt[n]{x})'=1/(n (\sqrt[n]{x})^{n-1}). It also can be written as (x^{1/n})'=(1/n)x^{(1/n) -1} .

## References

- M. Wahl, Derivatives Without Limits,
*Math Horizons*, Nov. 2008, p. 12, MAA - Jeff Suzuki, The Lost Calculus: Tangency and Optimization without Limits, Mathematics Magazine, Dec. 2005, p. 339-353, MAA
- Harold M. Edwards, Euler's definition of the derivative, Bull. Amer. Math. Soc. 44 (2007), 575-580, online at AMS.org
- Michael Livshits, You Could Simplify Calculus available online at arxiv.org
- Michael Livshits, Simplifying Calculus by Using Uniform Estimates, (13 slides for a 15 minutes talk) available online at mathfoolery.com