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For the infinite boomerang, we obtain:


[x2y2]=[x + y]  
x2(2yy') + y2(2x)=1 + y'  
y'(2x2y - 1)=1 - 2xy2  
y'=  

Therefore, at the point (0, 0), the slope of the graph is -1. Note that we cannot just plug any point we like into this formula--the point must be a solution to the original equation in order for the answer to make sense.

Differentiation of Inverse Functions

We can put the chain rule and implicit differentiation to work to find the derivative of an inverse function when we already know the derivative of the function itself. Suppose we are given a function f (x) with derivative f'(x) and let g(x) be its inverse, so that g(f (x)) = f (g(x)) = x. Differentiating both sides of f (g(x)) = x, we obtain:


f'(g(x))g'(x)=1  
g'(x)=  

Let us use this technique to find the derivative of the inverse sine function, f (x) = sin-1(x), defined on the interval [- 1, 1] and taking values in [- Π/2, Π/2]. Since f'(x) = cos(x), the formula tells us that g'(x) = 1/cos(sin-1(x)) = 1/. The derivatives of the other inverse trigonometric functions are as follows:


cos(x)=  
tan(x)=  

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