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Implicit
Differentiation

Learn to find derivatives of equations where \(y\) is not isolated: unlock slopes on circles, ellipses, and beyond.

AP Calculus BC

What Is Implicit Differentiation?

Most functions you have seen are written explicitly: \(y = f(x)\). But many important equations define \(y\) implicitly as a function of \(x\), without ever solving for \(y\).

Explicit
\(y = x^2 + 3x\)
Solve directly for \(y\).
Implicit
\(x^2 + y^2 = 25\)
Both \(x\) and \(y\) are intertwined.

Why Do We Need It?

Many curves cannot be expressed as a single function \(y = f(x)\). A circle, for example, requires two branches. Implicit differentiation handles the entire curve at once.

(6, 8) \(x^2 + y^2 = 100\) tangent x y

The Key Insight

The secret: treat \(y\) as a function of \(x\). Every time you differentiate a term containing \(y\), you must apply the chain rule and multiply by \(\frac{dy}{dx}\).

\(\frac{d}{dx}[x^2]\)
\(= 2x\)
No chain rule needed.
\(\frac{d}{dx}[y^2]\)
\(= 2y \cdot \frac{dy}{dx}\)
Chain rule applied.
\(\frac{d}{dx}[xy]\)
\(= y + x\frac{dy}{dx}\)
Product rule + chain.

The Formal Process

1
Differentiate both sides of the equation with respect to \(x\). Apply all standard rules (power, product, quotient, chain).
2
Attach \(\frac{dy}{dx}\) every time you differentiate a term that contains \(y\) (this is the chain rule).
3
Collect all \(\frac{dy}{dx}\) terms on one side of the equation and move everything else to the other side.
4
Factor out \(\frac{dy}{dx}\) and divide to isolate it.

Chain Rule in Action

\(F(x, y) = 0\) Differentiate w.r.t. x (chain rule) Solve for \(\frac{dy}{dx}\)

Explicit vs. Implicit

Explicit Implicit
Form \(y = f(x)\) \(F(x, y) = 0\)
Solve for \(y\)? Yes, already done Not always possible
Differentiation Direct: \(y' = f'(x)\) Chain rule on \(y\) terms
Result In terms of \(x\) only May include both \(x\) and \(y\)
Example \(y = \sqrt{25 - x^2}\) \(x^2 + y^2 = 25\)

How to Apply It

1
Write the equation so both sides are prepared for differentiation. Keep the equation balanced.
2
Differentiate every term with respect to \(x\). Use the power rule, product rule, and chain rule as needed.
3
Mark every \(y\)-term by multiplying by \(\frac{dy}{dx}\) (the chain rule factor).
4
Isolate \(\frac{dy}{dx}\): gather all \(\frac{dy}{dx}\) terms on one side, factor, and divide.
5
Simplify and, if needed, substitute a specific point \((x_0, y_0)\) to find the slope.

Example 1: The Circle

Find \(\frac{dy}{dx}\) for \(x^2 + y^2 = 25\).

1
Differentiate: \(\frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25]\)
2
Apply rules: \(2x + 2y\,\frac{dy}{dx} = 0\)
3
Isolate: \(2y\,\frac{dy}{dx} = -2x\)
4
Solve: \(\frac{dy}{dx} = -\frac{x}{y}\)

Example 2: Product Rule

Find \(\frac{dy}{dx}\) for \(xy + y^3 = 8\).

1
Differentiate: \(\frac{d}{dx}[xy] + \frac{d}{dx}[y^3] = 0\)
2
Product rule on \(xy\): \(y + x\,\frac{dy}{dx}\); chain rule on \(y^3\): \(3y^2\,\frac{dy}{dx}\)
3
Combine: \(y + x\,\frac{dy}{dx} + 3y^2\,\frac{dy}{dx} = 0\)
4
Factor: \(\frac{dy}{dx}(x + 3y^2) = -y\)
5
Result: \(\displaystyle\frac{dy}{dx} = \frac{-y}{x + 3y^2}\)

Example 3: Trigonometric Curves

Find \(\frac{dy}{dx}\) for \(\sin(x + y) = y^2\cos x\).

1
LHS: \(\cos(x+y)\cdot\bigl(1 + \frac{dy}{dx}\bigr)\) via chain rule.
2
RHS: \(2y\,\frac{dy}{dx}\cdot\cos x + y^2(-\sin x)\) via product rule.
3
Expand LHS: \(\cos(x+y) + \cos(x+y)\frac{dy}{dx}\).
4
Collect \(\frac{dy}{dx}\): \(\cos(x+y)\frac{dy}{dx} - 2y\cos x\,\frac{dy}{dx} = -y^2\sin x - \cos(x+y)\).
5
Solve: \(\displaystyle\frac{dy}{dx} = \frac{-y^2\sin x - \cos(x+y)}{\cos(x+y) - 2y\cos x}\).

Finding \(\frac{d^2y}{dx^2}\) Implicitly

You can find the second derivative implicitly by differentiating \(\frac{dy}{dx}\) again with respect to \(x\): treat \(\frac{dy}{dx}\) as a function of \(x\) and substitute back.

Connection to Related Rates

Implicit differentiation is the engine behind related rates problems. When both \(x\) and \(y\) change with time \(t\), you differentiate with respect to \(t\) instead of \(x\):

Implicit Diff.
Differentiate w.r.t. \(x\).
Find \(\frac{dy}{dx}\).
Related Rates
Differentiate w.r.t. \(t\).
Find \(\frac{dy}{dt}\) or \(\frac{dx}{dt}\).

Tangent Lines on Implicit Curves

To find the equation of a tangent line to an implicit curve at a point \((x_0, y_0)\):

1
Verify that the point satisfies the original equation.
2
Find \(\frac{dy}{dx}\) using implicit differentiation.
3
Evaluate \(\frac{dy}{dx}\) at \((x_0, y_0)\) to get the slope \(m\).
4
Write the tangent line: \(y - y_0 = m(x - x_0)\).

Common Mistakes

✕ Forgetting \(\frac{dy}{dx}\)
Differentiating \(y^2\) as \(2y\) instead of \(2y\,\frac{dy}{dx}\). The chain rule factor is essential.
✕ Missing Product Rule
Treating \(xy\) as a single variable. You must use the product rule: \(y + x\,\frac{dy}{dx}\).
✕ Algebra Errors
Incorrectly collecting \(\frac{dy}{dx}\) terms. Keep careful track of signs when moving terms across the equals sign.

Summary of Differentiation Rules

Expression \(\frac{d}{dx}\) Result Rule Used
\(x^n\) \(nx^{n-1}\) Power
\(y^n\) \(ny^{n-1}\,\frac{dy}{dx}\) Power + Chain
\(xy\) \(y + x\,\frac{dy}{dx}\) Product + Chain
\(\sin y\) \(\cos y \cdot \frac{dy}{dx}\) Trig + Chain
\(e^y\) \(e^y \cdot \frac{dy}{dx}\) Exponential + Chain
\(\ln y\) \(\frac{1}{y}\cdot\frac{dy}{dx}\) Log + Chain

Question 1 of 5

Given \(x^2 + y^2 = 49\), what is \(\frac{dy}{dx}\)?

Question 2 of 5

For the equation \(x^3 + y^3 = 6xy\), what is \(\frac{dy}{dx}\)?

Question 3 of 5

If \(\sin(xy) = x\), what is \(\frac{dy}{dx}\)?

Question 4 of 5

For \(e^y = x^2 + y\), find \(\frac{dy}{dx}\) and evaluate it at the point \((1, 0)\).

Question 5 of 5

For \(x^2 + y^2 = 25\), the second derivative \(\frac{d^2y}{dx^2}\) equals: