Implicit Differentiation and Inverse Functions

Not every curve arrives in the form $y=f(x)$. A circle, an energy constraint, or a relationship between two measured quantities may be written as one equation containing both $x$ and $y$. Such a relation can still have a tangent slope at a point. Implicit differentiation finds that slope by treating $y$ as a function of $x$ even when we do not solve the equation explicitly for $y$ first.

The key idea is the Chain Rule. If $y$ changes when $x$ changes, then differentiating $y^2$ with respect to $x$ gives $2y\frac{dy}{dx}$, not merely $2y$. The extra factor records the rate at which the dependent coordinate responds to the input. The same composition logic also explains derivatives of inverse functions: an inverse reverses a function’s input-output relationship, so its tangent slope is the reciprocal of the original nonzero slope.

By the end, you should be able to find a tangent slope from an implicit relation, identify the Chain Rule factor in every y-term, and use the inverse-function derivative formula with its conditions. You should also distinguish a relation’s local slope from an attempt to solve globally for all branches. This completes the derivative-construction unit and leads into derivative applications.

Differentiate y as a dependent quantity

Consider the circle

$$x^2+y^2=25.$$

Although this equation does not isolate $y$, it describes points on a circle of radius five. Differentiate both sides with respect to $x$. The $x^2$ term becomes $2x$. The $y^2$ term is a composite expression because $y=y(x)$, so the Chain Rule gives $2y\frac{dy}{dx}$. The derivative equation is

$$2x+2y\frac{dy}{dx}=0.$$

Now solve for the derivative:

$$\frac{dy}{dx}=-\frac{x}{y}.$$

At the point $(3,4)$, the slope is $-\frac34$. At $(-3,4)$, it is $\frac34$. Keeping $x$ and $y$ symbolic until after isolating $\frac{dy}{dx}$ creates a slope formula useful at every point with $y\ne0$. At points where $y=0$, the formula signals a vertical tangent rather than a finite slope.

Follow a stable workflow

Implicit differentiation is not a special trick. First differentiate every term with respect to $x$. Second, attach $\frac{dy}{dx}$ whenever differentiating an expression involving $y$. Third, use algebra to collect and solve for $\frac{dy}{dx}$. Only then substitute a point if the question asks for one particular tangent slope. This order reduces arithmetic clutter and makes the Chain Rule visible.

For $x^3+y^3=9xy$, differentiate term by term:

$$3x^2+3y^2\frac{dy}{dx} =9\left(x\frac{dy}{dx}+y\right).$$

The right side requires the Product Rule because both $x$ and $y$ change. Collect the terms containing $\frac{dy}{dx}$ on one side before solving. The goal is not to memorize a universal template but to recognize the ordinary product and chain structures already learned. Every factor has a reason.

Inverse functions reverse local slopes

If $f$ has an inverse, the composition $f^{-1}(f(x))=x$ returns each allowed input. Differentiate both sides using the Chain Rule:

$$(f^{-1})'(f(x))f'(x)=1.$$

Whenever $f'(x)\ne0$, solve for the inverse derivative:

$$(f^{-1})'(f(x))=\frac1{f'(x)}.$$

This formula is evaluated at matching points. If $f(a)=b$, then $(f^{-1})'(b)=\frac1{f'(a)}$. The original input $a$ becomes the inverse’s output, and the original output $b$ becomes the inverse’s input.

For $f(x)=x^3+1$, $f(2)=9$ and $f'(2)=12$. Therefore $(f^{-1})'(9)=\frac1{12}$. The answer is not $(f^{-1})'(2)$ because the inverse derivative must be evaluated at the original function’s output. This input-output swap is the main bookkeeping challenge in inverse-function problems.

The reciprocal formula requires a nonzero original derivative. A horizontal tangent in $f$ would become a vertical tangent in an inverse relation, so a finite reciprocal slope would not exist. The formula also presumes a local inverse branch; a function must pass a one-to-one test on the relevant domain before an inverse function can be discussed globally.

Bridge to applications

Implicit slopes make visible a broad modeling idea: constraints link rates. If a quantity is confined to a circle, a fixed-volume container, or a conservation relation, changing one coordinate forces changes in another. Differentiation converts the relation among quantities into a relation among local rates. In the applications unit, this will become the engine behind related-rates problems.

Before moving on, find $\frac{dy}{dx}$ for $x^2+xy+y^2=7$ and then evaluate it at any point you verify lies on the curve. Label the Product Rule term from differentiating $xy$ and the Chain Rule term from differentiating $y^2$. Finally, explain why $\frac{dy}{dx}$ at a point is not itself a point on the curve. It is a local slope, and that local linear behavior is the subject of the next lesson.

Unit 3 narrative challenge: the constrained camera arm

A camera mount traces the relation $x^2+y^2=25$, with coordinates measured in centimeters. At one instant the camera is at $(3\,\mathrm{cm},4\,\mathrm{cm})$, and its horizontal coordinate increases at $2\,\frac{\mathrm{cm}}{\mathrm{s}}$. The design team needs the vertical-coordinate rate at that instant to check whether the camera clears a housing. Write the constraint-rate equation before using the point values. Which sign should the vertical rate have, and why?

Solution and reasoning

Differentiate $x^2+y^2=25$ with respect to time: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Thus $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$. At the stated point, $\frac{dy}{dt}=-\frac34\left(2\,\frac{\mathrm{cm}}{\mathrm{s}}\right)=-1.5\,\frac{\mathrm{cm}}{\mathrm{s}}$. The rate is negative because on the upper branch of the circle, moving right from $(3,4)$ requires the height to decrease to preserve the fixed radius.