Implicit differentiation intuition

Some curves arrive as an equation that mixes $x$ and $y$, not as a formula $y=f(x)$. A circle $x^2+y^2=1$ is the usual first picture. You can solve for $y$ on the upper half, then differentiate. Implicit differentiation skips that solve. It treats $y$ as a function of $x$ that the equation already constrains, then differentiates both sides and solves for the slope.

The method looks like a trick until you hear the promise. If a smooth piece of the curve really is the graph of some $y(x)$, then every identity that holds along that piece remains an identity after you differentiate with respect to $x$. The derivative $y'(x)$ is the unknown slope. Algebra isolates it. This page is in the calculus cluster of the explainer library. The wider map is calculus foundations. The meaning of the slope you are about to isolate is what a derivative actually measures. A course shaped path that includes this page is the AP Calculus AB bridge.

What implicit is naming

Explicit means you are given $y$ as a recipe in $x$. Implicit means you are given a relation $F(x,y)=0$ and you trust that, near a point you care about, there is a unique smooth $y(x)$ hiding inside that relation. You do not need the recipe in hand. You need the relation, the chain rule, and a point where the curve is not vertical.

That last clause is the geometric warning. If the tangent is vertical, $y$ is not a function of $x$ in any neighborhood, because a vertical line meets the curve more than once, or the graph fails the vertical line test in every small window. You can still talk about $dx/dy$, or you can stay with a parametric description. Implicit differentiation with respect to $x$ is for pieces that pass the vertical line test locally.

The slope you get is still a best local rate. Once $y'(a)$ exists, the line $y=y(a)+y'(a)(x-a)$ is the best straight stand in for the curve near $x=a$. Implicit work is a way to find that number without first solving for $y$. It is not a different kind of derivative.

The chain rule does the work

Suppose $y=y(x)$ and you meet a term like $y^2$. As a function of $x$ that is the composition of $u\mapsto u^2$ with $y(x)$. The chain rule says

$$\frac{d}{dx}\bigl(y(x)^2\bigr)=2y(x)\,y'(x).$$

People write $2yy'$ and keep moving. The extra $y'$ is not decoration. It is the reminder that $y$ is riding on $x$. If you differentiate $y^2$ as if $y$ were a constant, you have changed the problem into a different curve, one where $y$ is frozen.

The same pattern hits a product, a sine of $y$, or an exponential of $xy$. Differentiate as usual, and every time a factor of $y$ is hit by the derivative you multiply by $y'$. Then collect every term that contains $y'$ on one side. The equation is linear in $y'$, even when it was nonlinear in $y$. That is why solving for the slope is usually easier than solving the original equation for $y$.

There is a compact way to see the same algebra. If $F(x,y(x))=0$ on an interval, differentiate with respect to $x$:

$$F_x+F_y\,y'=0.$$

Whenever $F_y\neq 0$,

$$y'=-\frac{F_x}{F_y}.$$

The long collection of terms you do by hand is this quotient, expanded. If you prefer to expand, do that. If you prefer the quotient, compute the two partial derivatives and divide. They must agree. A mismatch means an algebra slip, not a new slope.

A circle you can check by hand

Example Start with $F(x,y)=x^2+y^2-1$. Then $F_x=2x$ and $F_y=2y$, so $y'=-x/y$ when $y\neq 0$. The long way is the same: differentiate $x^2+y^2=1$ to get $2x+2yy'=0$, then divide.

Geometry agrees. On the unit circle the radius to $(x,y)$ is the vector $\langle x,y\rangle$. A tangent direction is perpendicular to that radius, so $\langle -y,x\rangle$ works. The run is the first component $-y$ and the rise is the second component $x$, so the slope is $x/(-y)=-x/y$ when $y\neq 0$. The implicit result matches the radius picture. You did not need a trigonometric parametrization, though $(\cos t,\sin t)$ gives the same slope $-\cos t/\sin t=-x/y$.

At $(0,1)$ you get $y'=0$, a horizontal tangent. At $(1,0)$ the formula $y'=-x/y$ blows up, which is honest. The tangent is vertical. Here $F_y=0$ and $F_x=2\neq 0$, so the curve is locally a graph of $x$ as a function of $y$ instead. The companion formula is $dx/dy=-F_y/F_x=0$ at that point. A horizontal tangent in the $(y,x)$ picture is a vertical line in the usual picture. Nothing failed. You asked for the wrong independent variable.

The upper semicircle $y=\sqrt{1-x^2}$ gives the same derivative by the ordinary chain rule: $y'=-x/\sqrt{1-x^2}=-x/y$ on $y>0$. Implicit work saved the square root, then agreed with it. The lower semicircle $y=-\sqrt{1-x^2}$ uses the same implicit formula, because $y$ is negative there and $-x/y$ changes sign in the way the lower tangent requires. One formula covers both branches. You still have to know which point you substituted.

A product, and a tangent line

Example Take $x^3+xy+y^3=3$ near $(1,1)$, which satisfies the equation because $1+1+1=3$. Differentiate, using the product rule on $xy$:

$$3x^2+\bigl(y+xy'\bigr)+3y^2 y'=0.$$

Collect $y'$ terms: $xy'+3y^2 y'=-3x^2-y$, so

$$y'=\frac{-3x^2-y}{x+3y^2}$$

when the denominator is not zero. At $(1,1)$ this is $-4/4=-1$. The tangent line is

$$y-1=-(x-1),$$

or $y=-x+2$. You have a best local line without solving a cubic for $y$.

The line is an approximation, not a replacement for the curve. At $x=1.1$ the line predicts $y=0.9$. The point on the curve at that $x$ solves $1.1^3+1.1y+y^3=3$, and the root near $0.9$ is about $0.886$. The line is high by about $0.014$. The step in $x$ was $0.1$, so the miss in $y$ is smaller than the step. The same gap shows up in the original equation. At $(1.1,0.9)$ the left side is $1.331+0.99+0.729=3.05$, near $3$ and not on the curve. If you needed the exact $y$ at $x=1.1$, you would still have to solve the cubic. The slope did not owe you that solve.

The quotient form gives the same fraction. With $F=x^3+xy+y^3-3$, one partial is $F_x=3x^2+y$ and the other is $F_y=x+3y^2$, so $y'=-(3x^2+y)/(x+3y^2)$. Same expression. Use whichever layout you can audit faster.

When both partials vanish

The implicit function theorem, in undergraduate language, says this. If $F$ is smooth and $F_y$ is not zero at a point on $F=0$, then near that point the level set is the graph of a unique smooth $y(x)$, and $y'=-F_x/F_y$. If $F_y=0$ and $F_x\neq 0$, switch the independent variable. If both partials vanish, stop. The level set may cross itself, pinch, or be an isolated point, and a single tangent may fail.

Example Let $y^2=x^3$, or $F=y^2-x^3$. Then $F_x=-3x^2$ and $F_y=2y$. Both are zero at the origin, which is on the curve. For $x>0$ there are two branches, $y=x^{3/2}$ and $y=-x^{3/2}$. On the upper branch the ordinary derivative is $(3/2)\sqrt{x}$, which approaches $0$ as $x$ approaches $0$ from the right. The implicit quotient $y'=3x^2/(2y)$ agrees with that on the upper branch, and it is undefined at the origin because you would be dividing by zero. The curve has a cusp there. Both branches come in with limiting slope $0$, and they do not form the graph of one function $y(x)$ on any two sided neighborhood of $x=0$, because $x$ cannot be negative. The algebra did not break. The promise "$y$ is a function of $x$ in a neighborhood" was never true at that point.

You do not need the full theorem to compute homework slopes. You do need the warning as a sanity check. Vertical tangents, crossings, and cusps are not bugs in your derivative rules. They are places where the story paused.

Related rates are the same idea

If $x$ and $y$ both depend on time $t$, and a constraint $F(x,y)=0$ holds for all $t$ in an interval, differentiate the constraint with respect to $t$. You obtain a linear relation between $x'(t)$ and $y'(t)$. That is implicit differentiation with a new independent variable. The geometry is unchanged. A moving point is required to stay on the curve, so its velocity is tangent to the curve.

Example Suppose a point stays on the circle $x^2+y^2=25$, and at the moment it passes through $(3,4)$ its horizontal rate is $x'=2$. Differentiate: $2xx'+2yy'=0$, so $y'=-xx'/y=-3\cdot 2/4=-3/2$ at that moment. The velocity vector is $\langle 2,-3/2\rangle$. It should be perpendicular to the radius $\langle 3,4\rangle$. The dot product is $3\cdot 2+4\cdot(-3/2)=6-6=0$. The rate you solved for really is tangent to the circle. If the dot product had not vanished, the differentiation or the arithmetic would be wrong. This is a model on a page, not a measurement from a lab.

The same habit checks an implicit solution of a differential equation. If a problem claims that a relation $F(x,y)=0$ solves $dy/dx=f(x,y)$, differentiate the relation and see whether the slope you get matches $f$ on the curve. The walkthrough for a first separated equation is separable differential equations. Bring this page with you when the solution is not written as $y$ equal to a formula.

What the method is not

• It is not a license to skip the point. A formula for $y'$ still needs a point on the curve before it becomes a number.
• It is not the slope of $F$. The gradient of $F$ is normal to the level set. The slope of the tangent line is a different object. They are tied together by $y'=-F_x/F_y$, which is exactly "the tangent is perpendicular to the gradient."
• It does not optimize under a constraint by itself. Finding a slope on a curve is not the same as finding a max of another function on that curve. That second job is Lagrange multipliers, and it uses the gradient picture you just met.
• A calculator that wants $y=f(x)$ cannot start the problem. It can audit a branch after you solve one.

How to know you understood

1. From $x^2+y^2=1$, derive $y'=-x/y$ and explain the vertical tangent at $(1,0)$ without calling it a broken formula.
2. Match that slope with the perpendicular to the radius, in one sentence.
3. At $(1,1)$ on $x^3+xy+y^3=3$, get slope $-1$ and write the tangent line. Say why $(1.1,0.9)$ is near the curve and not on it.
4. Give the cusp $y^2=x^3$ as a point where both partials vanish, and say what you refuse to do there.
5. On $x^2+y^2=25$ at $(3,4)$ with $x'=2$, get $y'=-3/2$ and check the dot product with the radius.

Check a branch on the numerical derivative calculator

The numerical derivative calculator needs an explicit formula in $x$. It is a comparison tool after you solve a branch, not a replacement for the implicit step. The numerical limit estimator can sample nearby values of that same explicit formula. Neither tool accepts $F(x,y)=0$.

The language of the limit under every difference quotient is epsilon delta intuition. You do not need epsilon and delta to finish a homework slope. You do need them if you want to say what "the slopes settle" means. Several variables, when you are ready, replace this one slope by a whole matrix of partials. That page is Jacobian matrices in plain language. The single quotient $y'=-F_x/F_y$ is the $1$ by $1$ preview.

Citations and further reading

• OpenStax, Calculus Volume 1, “Implicit Differentiation.” openstax.org
• Paul’s Online Math Notes, “Implicit Differentiation.” tutorial.math.lamar.edu
• MIT OpenCourseWare, 18.01SC Single Variable Calculus. ocw.mit.edu
• Stewart, J. Calculus, section on implicit differentiation, for drill problems after the idea is clear.