Derivatives of Inverse Functions

Derivatives of Inverse Functions

Let us assume that we have a function fox, and let us call it 3x – 1. We want to establish some sort of relationship between the slope of f and the slope of its inverse. To do this, we take these two functions and switch their x-variables and y-variables. This is all basic algebra.

Next, we solve for y by adding x to one side of the equation and dividing by 3. After that, we rewrite the equation in inverse form.

The slope of the resulting equation equals 1/3 and its inverse is written as 1/3 x plus 1/3.

Considering the slope of the line formed by plotting the points (1,3) and (3,1), what can we assume about inverse functions? Well, if the slope is three over one, it seems reasonable to assume that inverse functions have reciprocals. In fact, let's just sketch a graph of an inverse function that has a reciprocal slope.

Here is the graph of x squared minus 1. Its inverse is the square root of x.

We will not make this function because it is not defined at x equals negative one or zero. The graph looks like this. If we draw the tangent line at point a, b, then the corresponding point on this graph would be at (b, a).

The slopes of the tangent lines to a curve at two different points are reciprocals of one another. The derivative of a function is equal to the reciprocal of the function's slope at that point. If fis an inverse function of a function f, then the derivative off inverse is equal to m.

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The derivative of f(x) = bx is m when x = a.

Here, we have the big formula for inverse functions. The word to remember is reciprocal; it's basically 1 over another number "reciprocal" means 1 over.

Let f(x) be a function. If the slope of its inverse at b is 1/m, where m is the slope of fat a, then m and -1 are reciprocals. f(x)=3x-1 -> slope =3 y=3x-1 x=3y-1 x+1=3y x/3+1/3=y f^-1(x)=(1/3)x+1/3 -> slope = 1/3 Slope of Inverse? (xy)->(y,x).

Slopes of inverse are reciprocals [graph] Given f(a)=b and f^1(a) (f^-1)^1(b)=1/((f^1)(a)).

Reciprocal