To get derivatives of inverse trigonometric functions we were able to use implicit differentiation. Sometimes it is not possible/plausible to explicitly find inverse function, but we still want to find derivative of inverse function at certain point (slope). QUESTION: What is the relationship between derivatives of a function and its inverse ????

example: So

The relation But not you. (which you should know in the middle of the night).

A typical problem using this formula might look like this: Given:Find: example:

f -1 (a) = b. (f -1 )’(a) = tan . f’(b) = tan   +  = π/2

example: Find:

example: Find:

We have been more careful than usual in our statement of the differentiability result for inverse functions. You should notice that the differentiation formula for the inverse function involves division by f '(f -1 (x)). We must therefore assume that this value is not equal to zero. There is also a graphical explanation for this necessity. Example. The graphs of the cubing function f(x) = x 3 and its inverse (the cube root function) are shown below. Notice that f '(x)=3x 2 and so f '(0)=0. The cubing function has a horizontal tangent line at the origin. Taking cube roots we find that f -1 (0)=0 and so f '(f -1 (0))=0. The differentiation formula for f -1 can not be applied to the inverse of the cubing function at 0 since we can not divide by zero. This failure shows up graphically in the fact that the graph of the cube root function has a vertical tangent line (slope undefined) at the origin.