Implicit Differentiation and Derivatives of Inverse Functions

Sometimes a curve is described by a relationship between x and y without y being solved on its own. The circle x^2 + y^2 = 25 is a perfect example: y is mixed in with x, and there is no single neat formula for y. Implicit differentiation lets you find dy/dx anyway. The trick is to differentiate both sides with respect to x and treat y as a function of x. Every time you differentiate a y term, the chain rule tacks on a dy/dx. So the derivative of y^2 is 2y times dy/dx, and the derivative of xy needs the product rule plus a dy/dx. After differentiating, you collect the terms that contain dy/dx on one side and factor it out. Then you divide to solve for dy/dx. Once you have the derivative, you can plug in a point to find the gradient there, and even write the equation of the tangent line. This skill also unlocks derivatives of inverse functions, which is the next big idea.

A function and its inverse undo each other, so f(f inverse(x)) = x. Differentiate both sides of that equation with respect to x and the chain rule appears. On the left you get f'(f inverse(x)) times (f inverse)'(x), and on the right you get 1. Solving gives the famous formula (f inverse)'(x) = 1 divided by f'(f inverse(x)). In words: the slope of the inverse at a point is the reciprocal of the slope of the original function at the matching point. This one idea explains all six inverse trigonometric derivative formulas. For example, write y = arcsin x as sin y = x, differentiate implicitly to get cos y times dy/dx = 1, then use a trig identity to express cos y in terms of x. The result is dy/dx = 1/sqrt(1 - x^2). The same move works for arccos, arctan, and the rest.

Here are the three inverse trig derivatives you will use most often. The derivative of arcsin x is 1/sqrt(1 - x^2). The derivative of arccos x is -1/sqrt(1 - x^2), the same formula with a minus sign. The derivative of arctan x is 1/(1 + x^2). Each one comes from the same move: rewrite the inverse trig equation without the inverse, like sin y = x for arcsin, then differentiate implicitly and solve for dy/dx. Watch the chain rule when the inside is more than plain x. For arcsin(3x) you would multiply by 3, and for arctan(x^2) you multiply by 2x and replace x^2 inside the denominator. Memorize the three core formulas, and the chain rule handles the rest.

Free response questions about composite, implicit, and inverse functions reward clear setup more than speed. A typical question gives you a table of values for f, g, and their derivatives, then asks about a combination like f(g(x)) or the inverse of g. The winning habit is to write the general rule first, before touching any numbers. For a composite, write that the derivative of f(g(x)) is f'(g(x)) times g'(x). For an inverse, write (f inverse)'(x) = 1 divided by f'(f inverse(x)). Then read the table carefully: find the inside value first, look up its row, and only then multiply. If the question involves an implicit curve, differentiate term by term, gather every dy/dx on one side, and factor it out. Examiners give full credit for correct method even when arithmetic slips, so show each step in order.