For example, consider one of the following relationships between x and y : x – y – π = x + sin xy – y = In the first case, we can solve for y and rewrite the relationship as y = x – π . In the second case, it does not seem that there is an easy way to solve for y . Nevertheless, there is no doubt about the dependence of y on x in either of the cases. When a relationship between x and y is expressed in a way that it is easy to solve for y and write y = f ( x ), we say that y is given as an explicit function of x .
In the latter case it is implicit that y is a function of x and we say that the relationship of the second type, above, gives function implicitly . In this subsection, we learn to differentiate implicit functions. Example Find dy dx if x – y = π . Solution One way is to solve for y and rewrite the above as y = x – π But then dx = Alternatively , directly differentiating the relationship w.r.t., x , we have = d π Recall that d π means to differentiate the constant function taking value π everywhere w.r.t., x .
Thus = which implies that dx = Example Find dy , if y + sin y = cos x . Solution We differentiate the relationship directly with respect to x , i.e., (sin ) (cos ) which implies using chain rule = – sin x This gives −+ where y ≠ ( n + ) π . . Derivatives of inverse trigonometric functions We remark that inverse trigonometric functions are continuous functions, but we will not prove this.
Now we use chain rule to find derivatives of these functions. Example Find the derivative of f given by f ( x ) = sin – x assuming it exists. Solution Let y = sin – x . Then,