Lill’s method can be used to obtain the derivatives of polynomial equations in a graphical manner. The method to obtain the derivatives is described in the paper “NOTE ON LILL’S METHOD OF SOLUTION OF NUMERICAL EQUATIONS” by B. MEULENBELD. The paper complicates the matter by having Lill representations of polynomials that don’t have perpendicular segments. If the coefficients of the polynomial are real, then we should use perpendicular segments. Non-perpendicular segments are useful when representing polynomials with complex roots (see my paper Representing Polynomials with Complex Coefficients using Lill’s

Method).

In this post I want to show how to adapt the method described by MEULENBELD in order to obtain the first derivative of tan(θ), tan^2(θ), tan^3(θ) or other higher powers of tan(θ), for θ between -90 and 90 degrees.

The method to find the derivatives is not necessarily very practical, but nonetheless it shows again how Lill’s method can be connected to so many areas of math like trigonometry, algebra, geometry or calculus. I have a page dedicated to links related to Lill’s method.

### Derivative of tan(x)

The derivative of tan(θ)= sec^2(θ). To make things more simple ,we let θ=60 degrees. sec^2(60)=4. The figure below shows how to obtain sec^2(60).

In the image above P0B1 represents sec^2(θ). tan(P1P0A1)=tan(60)=P1A1=sqrt(3)=x. A1B1 is perpendicular to P0A1. Also A1P0B1=P1P0A1=60 degrees.

### Derivative of tan^2(θ)

The derivative of tan^2(θ)=2tan(θ)sec^2(θ). For θ=60 degrees, tan^2(60)=8sqrt(3). In the graph below, the derivative of tan^2(θ) is equal to (P0A1)( A2B1)=(2)(4sqrt(3))= 8sqrt(3).

In the image above P1A2=x^2=(sqrt(3))^2=3, where x=tan(60). The derivative of x^2 =2x=2sqrt(3)=(A2B1) /(P0A1). So we see that the derivative of x^2 is related to the derivative of tan^2(θ). Both involve the segments A2B1 and P0A1. We must divide A2B1 by P0A1 to obtain graphically the derivative of x^2, and we must multiply A2B1 by P0A1 to obtain the derivative of tan^2(θ) (where tan(θ)=x).

### Derivative of tan^3(θ)

The derivative of tan^3(θ)= 3tan^2(θ)sec^2(θ).For θ=60 degrees the value is 36. In the graph below, the derivative of tan^3(θ) is equal to (P0A1)(A3B2)=(2)(18)=36. It’s important to point out that A2A3 is perpendicular to A1A2 and B1B2 is perpendicular to A1B1.

P1A3=x^3=(sqrt(3))^3,where x=tan(60). The derivative of x^3=3x^2=18=(A3B2)/(P0B1). So we see again that the derivative of x^3 is related to the derivative of tan^3(θ), where x=tan(θ).

### Derivative of Higher Powers of tan(θ)

For higher powers of tan(θ) we need to construct additional perpendicular segments. The segment AnAn-1 is always perpendicular to An-1An-2 and BnBn-1 is perpendicular to Bn-1Bn-2. The point An should always be on the x-axis or the y-axis. The point Bn-1 should always be on the line passing through the points An and An-1.

The derivative of tan^n(θ) is equal to (P0A1)(AnBn-1), for n>1. The derivative of x^n is equal to (AnBn-1)/(P0A1), where x=tan(θ). The case of n=1 is special , since the derivative of tan(θ) is equal to P0B1.

### Final Notes

The method can probably be adapted to find the second order derivative or higher order derivatives of the powers of tan(θ). However, the complexity would increase as you deal with even more points and segments.

There is also the problem of determining the sign of the value of the derivative. The segments gives us the absolute value. Rules for determining when the segments represent positive or negative values can be established, but I consider that a more minor issue. The intent of this post is to show a general outline of the method.