Now the binomial expansion is only defined for integers, therefore this proof is only for integer exponents of x. We know that the equation of the straight line that passes through the point (x y y It is noted that the slope of the tangent line to the curve f(x)y at. Now each term of the denominator contains an h but all contain higher powers of h than 1 except for the first term, therefore when we divide each term by h and take the limit as h → 0, all terms but the first vanish, leaving Here is the derivative expression for this function it's just a matter of plugging the function into the difference quotient expression: The function (see graph below right) has a positive slope for all x in its domain, except for x = 0, at which the slope is undefined. Your line is f(x) x which can be written y. It has a vertical asymptote at x = 1, therefore finding the slope of f(x) at x = 1 makes no sense – the function has no actual value there. The slope-intercept form of the equation of a line is y mx + b, where m is the slope and b is the y-intercept. This is an interesting function–a rational function. We'll wait for a while to actually prove that this limit is the slope of f(x) at x.

The graphs below will help you visualize how making the distance between our two points smaller gives us a better and better estimate of the slope of f(x) at x.

That is the essence of differential calculus: Finding the limit of a slope function as the change in the independent variable approaches zero. Now do the following thought exercise: Imagine that Δx gets smaller and smaller, and eventually vanishes or becomes 'infinitessimally small.' In the limit that Δx = 0, we would have the exact slope of our function at point x. This abstract concept has a variety of concrete realizations. Convince yourself that this equation still just represents rise over run, Δy / Δx (and don't forget that y = f(x)). Finding the slope of a curve at a point is one of two fundamental problems in calculus. The slope of our secant line ( magenta), m, is written above the graph in terms of x, Δx, f(x) and f(x+Δx).

This will allow us to reduce the width of the interval between them (the x-distance) by reducing Δx. To do that, we'll first change our notation a little and use x and ( x + Δx) instead of x 1 and x 2 to label our two points. This time we'll try to find the slope of our curve at some specific point x.

Find the slope of fx equation