• Differentiate equations using implicit differentiation

Consider the circle . Find the slope of the tangent line passing through the point .

From geometry we know that all lines tangent to a circle are perpendicular to a radius. We know that the slope of the line passing through the points and is 1. So the slope of the line tangent to the circle at is -1.

We can confirm this result with differential calculus. Take the equation of the circle and solve for y.

Since the point of tangency is we select the positive root and find its derivative.


So the slope of the tangent line at is given by .

is not a function, however we can solve the equation for y and get . We say and are implied functions of x. We can differentiate the equation directly term by term as long as we realize y is a function of x and use the chain rule appropriately.

On the first term on the left we use the simple power rule. On the second term on the left we use the generalized power rule recognizing y as an implied function of x. We indicate the derivative of y with respect to x with the symbol .

Now we solve this equation algebraically for the symbol .

So the slope of the tangen line passing through is given by

Example:
Find implicitly for the equation .
Solution: