# The Slope of an Elliptic Curve

*Before we continue with our investigation of elliptic curves in a finite field, we need to clarify some important calculations, that will be used for adding and doubling values in a finite field, on an elliptic curve.*

Finding the gradient (slope) of the tangent line at a point on an elliptic curve given in Weierstrass form involves using partial derivatives. The standard Weierstrass form of an elliptic curve is: *y^2 = x^3 + Ax + B* where *A, B* are constants in *F* , the field across which *E*, the elliptic curve, is defined.

To find the gradient at a point on this curve, you’ll need to differentiate implicitly since the equation involves both x and y interdependently.

Implicit Function - given the equation:

### Differentiate implicitly with respect to *x*

Differentiate both sides of the equation with respect to *x*. Using implicit differentiation, the derivative of *y^2* with respect to *x* is:

(applying the chain rule) and the derivative of:

with respect to x is:

Setting the derivatives equal gives:

To find dy/dx (the gradient of the tangent line at any point on the curve), rearrange the equation:

This equation gives you the slope of the tangent line at any point (x, y) on an elliptic curve in the Weierstrass form, provided that:

since dividing by zero is undefined.

Note that, in the form of the derivative where only *x* is used, the equation yields two solutions, precisely because there are 2 points, each reflected across the x-axis, for any value of *x* on the curve, each with a gradient *g_n*, such that *g_1 = −g_2*.

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