The Central Idea
Ordinary differentiation often begins with an explicit function such as
\[ y=x^3-2x. \]Then \(y\) is already isolated, so we differentiate directly:
\[ \frac{dy}{dx}=3x^2-2. \]But a curve may instead be given implicitly,
\[ x^2+y^2=25, \]or parametrically,
\[ x=t^2+1, \qquad y=t^3-t. \]In both cases, the curve still has tangent slopes. We simply need a different route to calculate them.
Explicit, Implicit, and Parametric Descriptions
An explicit description isolates one variable as a function of another:
\[ y=f(x). \]An implicit description places the variables together in a relation:
\[ F(x,y)=0. \]A parametric description defines both coordinates in terms of a parameter:
\[ x=x(t), \qquad y=y(t). \]These are different representations of curves, not different kinds of geometry. The same geometric curve can sometimes be represented in more than one of these forms.
Why Implicit Differentiation Is Needed
Consider the circle
\[ x^2+y^2=25. \]Solving for \(y\) gives two branches:
\[ y=\pm\sqrt{25-x^2}. \]Working with the original relation is often cleaner. Implicit differentiation lets us differentiate the equation directly and obtain one slope formula that applies wherever the relevant derivative exists.
Treat \(y\) as a Function of \(x\)
The key idea in implicit differentiation is that \(y\) depends on \(x\), even when that dependence has not been written explicitly.
Conceptually, we may think
\[ y=y(x). \]Therefore, whenever we differentiate an expression involving \(y\) with respect to \(x\), the chain rule appears.
For example,
\[ \frac{d}{dx}(y^2) =2y\frac{dy}{dx}. \]The extra factor \(dy/dx\) is essential.
The Basic Implicit-Differentiation Procedure
- Differentiate both sides of the equation with respect to \(x\).
- Apply ordinary derivative rules to expressions involving \(x\).
- Apply the chain rule to expressions involving \(y\).
- Collect all terms containing \(dy/dx\).
- Factor out \(dy/dx\) when necessary.
- Solve algebraically for \(dy/dx\).
- Simplify the result when useful.
Example: A Circle
Differentiate
\[ x^2+y^2=25. \]Differentiate both sides with respect to \(x\):
\[ 2x+2y\frac{dy}{dx}=0. \]Move \(2x\) to the other side:
\[ 2y\frac{dy}{dx}=-2x. \]Divide by \(2y\):
\[ \boxed{\frac{dy}{dx}=-\frac{x}{y}}. \]Why \(d(y^2)/dx\) Is Not Just \(2y\)
If \(y\) were the independent variable, then
\[ \frac{d}{dy}(y^2)=2y. \]But implicit differentiation uses differentiation with respect to \(x\). Since \(y=y(x)\), we apply the chain rule:
\[ \frac{d}{dx}[y(x)]^2 =2y(x)y'(x). \]Thus,
\[ \boxed{\frac{d}{dx}(y^2)=2y\frac{dy}{dx}}. \]Differentiating Powers of \(y\)
More generally,
\[ \boxed{\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}}. \]For example,
\[ \frac{d}{dx}(y^5) =5y^4\frac{dy}{dx}. \]This is simply the power rule combined with the chain rule.
Example: Mixed Powers of \(x\) and \(y\)
Find \(dy/dx\) if
\[ x^3+y^3=6xy. \]Differentiate both sides:
\[ 3x^2+3y^2\frac{dy}{dx} =6\left(y+x\frac{dy}{dx}\right). \]The right side requires the product rule because both \(x\) and \(y\) change with \(x\).
Expand:
\[ 3x^2+3y^2\frac{dy}{dx} =6y+6x\frac{dy}{dx}. \]Collect derivative terms on one side:
\[ 3y^2\frac{dy}{dx}-6x\frac{dy}{dx} =6y-3x^2. \]Factor:
\[ \frac{dy}{dx}(3y^2-6x)=6y-3x^2. \]Therefore,
\[ \boxed{ \frac{dy}{dx} =\frac{2y-x^2}{y^2-2x} }. \]Products Containing \(y\)
If a term contains both \(x\) and \(y\), such as
\[ xy, \]use the product rule:
\[ \frac{d}{dx}(xy) =1\cdot y+x\frac{dy}{dx}. \]Thus,
\[ \boxed{\frac{d}{dx}(xy)=y+x\frac{dy}{dx}}. \]Example: Product Terms
Differentiate
\[ x^2+xy+y^2=7. \]Term by term,
\[ 2x+ \left(y+x\frac{dy}{dx}\right) +2y\frac{dy}{dx}=0. \]Collect the derivative terms:
\[ x\frac{dy}{dx}+2y\frac{dy}{dx} =-2x-y. \]Factor and solve:
\[ (x+2y)\frac{dy}{dx}=-2x-y, \] \[ \boxed{ \frac{dy}{dx} =-\frac{2x+y}{x+2y} }. \]Functions of \(y\) Also Need the Chain Rule
The principle extends beyond powers. If \(G\) is differentiable, then
\[ \frac{d}{dx}G(y) =G'(y)\frac{dy}{dx}. \]For example,
\[ \frac{d}{dx}(\sin y) =\cos y\frac{dy}{dx}, \]and
\[ \frac{d}{dx}(e^y) =e^y\frac{dy}{dx}. \]Example: A Trigonometric Relation
Suppose
\[ \sin y=x^2. \]Differentiate both sides:
\[ \cos y\frac{dy}{dx}=2x. \]Therefore,
\[ \boxed{ \frac{dy}{dx}=\frac{2x}{\cos y} }. \]The derivative can naturally remain in terms of both \(x\) and \(y\).
Evaluating an Implicit Derivative at a Point
For the circle
\[ x^2+y^2=25, \]we found
\[ \frac{dy}{dx}=-\frac{x}{y}. \]At the point \((3,4)\),
\[ \left.\frac{dy}{dx}\right|_{(3,4)} =-\frac{3}{4}. \]So the tangent slope there is
\[ \boxed{-\frac{3}{4}}. \]Tangent Lines from Implicit Differentiation
Once the slope is known, use point-slope form.
At \((3,4)\), the circle has slope \(-3/4\), so its tangent line is
\[ y-4=-\frac{3}{4}(x-3). \]This can be left in point-slope form or rearranged as needed.
Normal Lines from Implicit Differentiation
If a tangent line has nonzero finite slope \(m\), a perpendicular normal line has slope
\[ -\frac{1}{m}. \]For the circle at \((3,4)\), the tangent slope is \(-3/4\), so the normal slope is
\[ \frac{4}{3}. \]Thus the normal line is
\[ y-4=\frac{4}{3}(x-3). \]A Compact Formula for \(F(x,y)=0\)
Suppose a curve is given by
\[ F(x,y)=0. \]Differentiating implicitly gives
\[ F_x+F_y\frac{dy}{dx}=0, \]where \(F_x\) and \(F_y\) denote the partial derivatives of \(F\) with respect to its first and second variables.
When \(F_y\neq0\),
\[ \boxed{ \frac{dy}{dx}=-\frac{F_x}{F_y} }. \]This is a compact advanced form of the same implicit-differentiation procedure. The underlying method does not change.
When Solving for \(y\) First Is Reasonable
Implicit differentiation is a tool, not a requirement. If an equation can be solved easily for \(y\), explicit differentiation may be simpler.
For example,
\[ 2x+y=7 \]gives immediately
\[ y=7-2x, \]so
\[ \frac{dy}{dx}=-2. \]Implicit differentiation gives the same result, but isolating \(y\) first is faster here.
Why Parametric Curves Are Useful
Some curves are naturally described by motion. Instead of saying directly how \(y\) depends on \(x\), we say how both coordinates depend on a parameter \(t\):
\[ x=x(t), \qquad y=y(t). \]The parameter often represents time, but it can represent any quantity that organizes the motion along the curve.
A Simple Parametric Example
Consider
\[ x=t, \qquad y=t^2. \]Eliminating the parameter gives
\[ y=x^2. \]So this parametric system traces the familiar parabola. The parameter tells us which point is reached for each value of \(t\).
The Parametric Derivative Formula
If both coordinates depend on \(t\), then the chain rule relates their rates:
\[ \frac{dy}{dt} =\frac{dy}{dx}\frac{dx}{dt}. \]When \(dx/dt\neq0\), solve for \(dy/dx\):
\[ \boxed{ \frac{dy}{dx} =\frac{dy/dt}{dx/dt} }. \]In words, the slope of the curve is the vertical rate divided by the horizontal rate.
Why the Ratio Makes Sense
Suppose a moving point has horizontal velocity
\[ \frac{dx}{dt}=4 \]and vertical velocity
\[ \frac{dy}{dt}=6. \]Then for each unit of horizontal change, the vertical change occurs at the ratio
\[ \frac{6}{4}=\frac{3}{2}. \]That ratio is precisely the tangent slope:
\[ \frac{dy}{dx}=\frac{3}{2}. \]Example: Differentiate a Parametric Curve
Let
\[ x=t^2+1, \qquad y=t^3-3t. \]Differentiate each coordinate with respect to \(t\):
\[ \frac{dx}{dt}=2t, \] \[ \frac{dy}{dt}=3t^2-3. \]Therefore, when \(t\neq0\),
\[ \boxed{ \frac{dy}{dx} =\frac{3t^2-3}{2t} }. \]Evaluating a Parametric Slope
For
\[ x=t^2+1, \qquad y=t^3-3t, \]the slope is
\[ \frac{dy}{dx} =\frac{3t^2-3}{2t}. \]At \(t=2\),
\[ \frac{dy}{dx} =\frac{12-3}{4} =\frac{9}{4}. \]The point at \(t=2\) is
\[ x=5, \qquad y=2. \]So the curve has slope \(9/4\) at \((5,2)\).
Tangent Lines to Parametric Curves
Once the parameter value is known, calculate three things:
- the point \((x(t),y(t))\);
- the horizontal rate \(dx/dt\);
- the vertical rate \(dy/dt\).
Then form \(dy/dx\) and use point-slope form.
In the preceding example at \(t=2\), the tangent line is
\[ \boxed{ y-2=\frac{9}{4}(x-5) }. \]Parametric Curves Can Trace Shapes Difficult to Express as \(y=f(x)\)
A circle can be parameterized as
\[ x=r\cos t, \qquad y=r\sin t. \]Differentiate:
\[ \frac{dx}{dt}=-r\sin t, \qquad \frac{dy}{dt}=r\cos t. \]Therefore, when \(\sin t\neq0\),
\[ \frac{dy}{dx} =\frac{r\cos t}{-r\sin t} =-\cot t. \]The parameterization represents the entire circle without splitting it into upper and lower explicit branches.
Implicit and Parametric Slopes Agree
For the circle
\[ x^2+y^2=r^2, \]implicit differentiation gives
\[ \frac{dy}{dx}=-\frac{x}{y}. \]Under the parameterization
\[ x=r\cos t, \qquad y=r\sin t, \]this becomes
\[ -\frac{x}{y} =-\frac{r\cos t}{r\sin t} =-\cot t, \]exactly matching the parametric derivative.
Horizontal Tangents for Parametric Curves
A horizontal tangent generally occurs when the vertical rate is zero while the horizontal rate is nonzero:
\[ \boxed{ \frac{dy}{dt}=0, \qquad \frac{dx}{dt}\neq0 }. \]Then
\[ \frac{dy}{dx}=0. \]Vertical Tangents for Parametric Curves
A vertical tangent often occurs when the horizontal rate is zero while the vertical rate is nonzero:
\[ \boxed{ \frac{dx}{dt}=0, \qquad \frac{dy}{dt}\neq0 }. \]In that situation, the ratio
\[ \frac{dy/dt}{dx/dt} \]has a zero denominator, corresponding to a vertical tangent direction.
Example: Horizontal Tangents
Let
\[ x=t^2+1, \qquad y=t^3-3t. \]We have
\[ \frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2-3. \]Set the vertical rate equal to zero:
\[ 3t^2-3=0. \]Thus
\[ t=\pm1. \]Since \(dx/dt=\pm2\neq0\) at those values, both produce horizontal tangents.
Example: A Vertical Tangent
For the same curve,
\[ \frac{dx}{dt}=2t. \]This equals zero at
\[ t=0. \]Meanwhile,
\[ \left.\frac{dy}{dt}\right|_{t=0}=-3\neq0. \]Therefore the curve has a vertical tangent at \(t=0\). The point is
\[ (x,y)=(1,0). \]When Both Parametric Rates Are Zero
If
\[ \frac{dx}{dt}=0 \qquad\text{and}\qquad \frac{dy}{dt}=0 \]at the same parameter value, the simple ratio gives the indeterminate form \(0/0\).
We cannot automatically classify the tangent as horizontal or vertical. The local behavior must be examined more carefully, often by simplifying the ratio, using limits, or studying the parameterization near that value.
The Parameter Is Not Necessarily \(x\)
In a parametric curve,
\[ x=x(t), \qquad y=y(t), \]the parameter \(t\) organizes the curve. It may represent time, angle, distance traveled, or another quantity.
The formula
\[ \frac{dy}{dx} =\frac{dy/dt}{dx/dt} \]converts rates with respect to the parameter into a rate of \(y\) with respect to \(x\).
Orientation Matters in a Parametric Description
A parametric system tells us not only which points lie on the curve, but also the order in which those points are traced as the parameter changes.
For example,
\[ x=\cos t, \qquad y=\sin t \]traces the unit circle counterclockwise as \(t\) increases from \(0\) to \(2\pi\).
A different parameterization can trace the same geometric curve at a different speed or in the opposite direction.
Eliminating the Parameter Can Be Useful, but It Is Not Required
Sometimes we can eliminate \(t\) to recover a Cartesian equation. For example,
\[ x=2t+1, \qquad y=t^2 \]gives
\[ t=\frac{x-1}{2}, \]so
\[ y=\left(\frac{x-1}{2}\right)^2. \]But parametric differentiation can find the slope directly without this conversion.
Example: Direct Parametric Differentiation
For
\[ x=2t+1, \qquad y=t^2, \]differentiate:
\[ \frac{dx}{dt}=2, \qquad \frac{dy}{dt}=2t. \]Therefore,
\[ \boxed{ \frac{dy}{dx}=t }. \]Since \(t=(x-1)/2\), this agrees with differentiating the explicit form.
Implicit Differentiation as a Rate Relationship
Implicit differentiation does more than produce tangent slopes. It reveals how variables constrained by an equation must change together.
If
\[ x^2+y^2=25, \]then
\[ 2x+2y\frac{dy}{dx}=0. \]This equation states that changes in \(x\) and \(y\) cannot occur independently if the point must remain on the circle.
Parametric Differentiation as a Rate Ratio
Parametric differentiation has an equally useful interpretation. If a moving point has coordinate rates
\[ \frac{dx}{dt} \qquad\text{and}\qquad \frac{dy}{dt}, \]then the geometric slope along its path is obtained by dividing the vertical rate by the horizontal rate:
\[ \boxed{ \frac{dy}{dx} =\frac{dy/dt}{dx/dt} }. \]This connects motion, coordinate rates, and tangent geometry.
Choosing Between the Methods
Use implicit differentiation when the variables are tied together by an equation such as
\[ F(x,y)=0. \]Use parametric differentiation when both coordinates are given in terms of a parameter:
\[ x=x(t), \qquad y=y(t). \]If the curve has an easy explicit form \(y=f(x)\), ordinary differentiation may still be the most efficient choice.
Common Error: Forgetting \(dy/dx\)
From
\[ x^2+y^2=25, \]an incorrect step is
\[ 2x+2y=0. \]The derivative of \(y^2\) with respect to \(x\) must include the chain-rule factor:
\[ \boxed{ 2x+2y\frac{dy}{dx}=0 }. \]Common Error: Applying the Product Rule Incorrectly to \(xy\)
Because both \(x\) and \(y\) vary with \(x\),
\[ \frac{d}{dx}(xy) \]is not merely \(y\) and is not merely \(x\,dy/dx\). The product rule gives
\[ \boxed{ \frac{d}{dx}(xy) =y+x\frac{dy}{dx} }. \]Common Error: Solving for \(dy/dx\) Too Early
During implicit differentiation, it is usually safer to differentiate the entire equation first and only then collect the derivative terms.
Trying to isolate \(dy/dx\) before all product-rule and chain-rule terms have been expanded often causes missing terms or sign errors.
Common Error: Inverting the Parametric Ratio
The correct formula is
\[ \boxed{ \frac{dy}{dx} =\frac{dy/dt}{dx/dt} }. \]The numerator must correspond to the variable in the numerator of \(dy/dx\), and the denominator must correspond to the variable in the denominator.
Common Error: Cancelling the \(dt\) Symbols as Ordinary Fractions
The notation
\[ \frac{dy/dt}{dx/dt} \]resembles fraction cancellation, but the formula is justified by the chain rule, not by treating differentials as ordinary independent numbers.
The notation is suggestive and useful, but the derivative relationship is the mathematical reason the formula works.
Common Error: Declaring a Vertical Tangent Whenever \(dx/dt=0\)
A zero horizontal rate is strong evidence for a vertical tangent only when the vertical rate is nonzero:
\[ \frac{dx}{dt}=0, \qquad \frac{dy}{dt}\neq0. \]If both rates vanish, the local behavior requires additional analysis.
A Reliable Implicit-Differentiation Workflow
- Identify the equation relating \(x\) and \(y\).
- Differentiate every term with respect to \(x\).
- Attach \(dy/dx\) whenever differentiating through a \(y\)-dependent expression.
- Use product, quotient, and chain rules wherever the algebraic structure requires them.
- Move all \(dy/dx\) terms to one side.
- Move the remaining terms to the other side.
- Factor out \(dy/dx\).
- Divide to isolate \(dy/dx\).
- If a point is given, substitute its coordinates only after deriving the slope formula unless earlier substitution clearly simplifies the work.
A Reliable Parametric-Differentiation Workflow
- Differentiate \(x(t)\) to find \(dx/dt\).
- Differentiate \(y(t)\) to find \(dy/dt\).
- Form \((dy/dt)/(dx/dt)\).
- Simplify where \(dx/dt\neq0\).
- If a specific parameter value is given, find the corresponding point.
- Evaluate the slope at that parameter value.
- Use point-slope form if a tangent line is required.
- Check separately for horizontal or vertical tangents when either coordinate rate is zero.
Example: Full Implicit Workflow
Find the slope of
\[ x^2+3xy+2y^2=12 \]at a point on the curve.
Differentiate:
\[ 2x+3\left(y+x\frac{dy}{dx}\right) +4y\frac{dy}{dx}=0. \]Expand:
\[ 2x+3y+3x\frac{dy}{dx} +4y\frac{dy}{dx}=0. \]Collect derivative terms:
\[ (3x+4y)\frac{dy}{dx}=-(2x+3y). \]Therefore,
\[ \boxed{ \frac{dy}{dx} =-\frac{2x+3y}{3x+4y} }. \]A particular slope is found by substituting the coordinates of the chosen point into this formula.
Example: Full Parametric Workflow
Suppose
\[ x=t^2-2t, \qquad y=t^3-3t. \]Differentiate:
\[ \frac{dx}{dt}=2t-2, \qquad \frac{dy}{dt}=3t^2-3. \]Therefore,
\[ \frac{dy}{dx} =\frac{3t^2-3}{2t-2}. \]Factor:
\[ \frac{dy}{dx} =\frac{3(t-1)(t+1)}{2(t-1)}. \]For \(t\neq1\), this simplifies to
\[ \boxed{ \frac{dy}{dx}=\frac{3}{2}(t+1) }. \]At \(t=1\), however, both coordinate derivatives vanish, so that parameter value must be examined separately rather than handled by ordinary cancellation alone.
Practice 1: Basic Implicit Differentiation
Find \(dy/dx\):
\[ x^2+y^2=16. \]Practice 2: Powers of \(y\)
Find \(dy/dx\):
\[ x^3+y^4=10. \]Practice 3: Product Term
Find \(dy/dx\):
\[ x^2+xy=6. \]Practice 4: Tangent Slope
Find the slope of the curve
\[ x^2+y^2=25 \]at \((4,3)\).
Practice 5: Trigonometric Implicit Differentiation
Find \(dy/dx\):
\[ x+\sin y=2. \]Practice 6: Basic Parametric Derivative
Find \(dy/dx\):
\[ x=t^2, \qquad y=t^3+1. \]Practice 7: Parametric Slope at a Value
For
\[ x=3t+1, \qquad y=t^2-2, \]find the slope when \(t=3\).
Practice 8: Horizontal Tangent
For
\[ x=t^2+2t, \qquad y=t^2-4t, \]find the parameter value at which the tangent is horizontal, provided \(dx/dt\neq0\).
Practice 9: Vertical Tangent
For
\[ x=t^2-4t, \qquad y=t^2+t, \]find the parameter value at which the tangent is vertical, provided \(dy/dt\neq0\).
Practice 10: Identify the Error
A student differentiates
\[ x^2+y^3=7 \]and writes
\[ 2x+3y^2=0. \]Explain the error and correct the derivative equation.
Practice Answers
Practice 1
\[ 2x+2y\frac{dy}{dx}=0, \]so
\[ \boxed{ \frac{dy}{dx}=-\frac{x}{y} }. \]Practice 2
\[ 3x^2+4y^3\frac{dy}{dx}=0, \]so
\[ \boxed{ \frac{dy}{dx}=-\frac{3x^2}{4y^3} }. \]Practice 3
\[ 2x+y+x\frac{dy}{dx}=0. \]Therefore,
\[ \boxed{ \frac{dy}{dx}=-\frac{2x+y}{x} }. \]Practice 4
For the circle,
\[ \frac{dy}{dx}=-\frac{x}{y}. \]At \((4,3)\),
\[ \boxed{-\frac{4}{3}}. \]Practice 5
\[ 1+\cos y\frac{dy}{dx}=0, \]so
\[ \boxed{ \frac{dy}{dx}=-\frac{1}{\cos y} }. \]Practice 6
\[ \frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2. \]Thus, for \(t\neq0\),
\[ \boxed{ \frac{dy}{dx}=\frac{3t}{2} }. \]Practice 7
\[ \frac{dx}{dt}=3, \qquad \frac{dy}{dt}=2t. \]At \(t=3\),
\[ \boxed{ \frac{dy}{dx}=2 }. \]Practice 8
\[ \frac{dx}{dt}=2t+2, \qquad \frac{dy}{dt}=2t-4. \]A horizontal tangent requires \(dy/dt=0\), so
\[ 2t-4=0 \quad\Rightarrow\quad \boxed{t=2}. \]At \(t=2\), \(dx/dt=6\neq0\).
Practice 9
\[ \frac{dx}{dt}=2t-4, \qquad \frac{dy}{dt}=2t+1. \]A vertical tangent requires \(dx/dt=0\), so
\[ 2t-4=0 \quad\Rightarrow\quad \boxed{t=2}. \]At \(t=2\), \(dy/dt=5\neq0\).
Practice 10
The student forgot that \(y\) depends on \(x\). The chain rule gives
\[ \boxed{ 2x+3y^2\frac{dy}{dx}=0 }. \]Therefore,
\[ \boxed{ \frac{dy}{dx}=-\frac{2x}{3y^2} }. \]What to Remember
For implicit differentiation, remember that \(y\) is a function of \(x\). Therefore, differentiating a \(y\)-expression introduces \(dy/dx\):
\[ \boxed{ \frac{d}{dx}(y^n) =ny^{n-1}\frac{dy}{dx} }. \]For parametric differentiation,
\[ \boxed{ \frac{dy}{dx} =\frac{dy/dt}{dx/dt} } \]whenever \(dx/dt\neq0\).
- implicit differentiation works directly with relations involving \(x\) and \(y\);
- the chain rule is the mechanism behind every \(dy/dx\) factor;
- products involving \(y\) still require the product rule;
- parametric differentiation compares vertical and horizontal rates;
- horizontal and vertical tangents require separate checks of the coordinate rates;
- when both parametric rates vanish, additional local analysis is required.
Where We Go Next
So far, differentiation has focused mainly on the first derivative: the local rate of change or tangent slope. But a derivative is itself a function, and that function can often be differentiated again.
The next lesson, Higher Derivatives, develops second, third, and higher-order derivatives and explains how repeated differentiation captures acceleration, curvature behavior, changing sensitivity, and other layers of change.