Differentiate both sides: Preserve equality under differentiation

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | Related Principles | FAQ | How This Fits

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The Principle

The move: Apply $\frac{d}{dx}$ to both sides of an identity or implicit relation $u(x)=v(x)$, producing a new equation between the derivatives.

The invariant: This preserves equality under differentiation, so the transformed equation remains true wherever the relation holds and both sides are differentiable.

Pattern: $u(x)=v(x) \quad\longrightarrow\quad u'(x)=v'(x)$

Legal ✓	Illegal ✗
$x^2 + y^2 = 25 \Rightarrow 2x + 2y\,y' = 0$; the relation holds and both sides are differentiable with respect to $x$	$x = x^2$ at $x=1 \not\Rightarrow 1 = 2x$; the equation is true only at one point, not as an identity or implicit relation in a neighborhood where the move can act

The tempting mistake is to treat any equation that happens to be true at one point as automatically eligible. It is not. The move acts on a differentiable relation, not on a one-point coincidence or a random equation-to-solve.

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Conditions of Applicability

Condition: identity or implicit relation; u and v differentiable

Differentiability is not enough. The equation must function as an identity or implicit relation in the setting where you are differentiating, not merely be true at one point, and each side must define a differentiable function of the same variable in the part of the domain where you are working.

Before applying, check: confirm that the equation is being used as an identity or implicit relation, and that both sides have derivatives in the current setting.

• If the equation is only true at one isolated point, differentiating both sides can create a false new equation; the move is not valid there.
• If one side has a corner, cusp, vertical tangent, or domain break where you want to differentiate, the move is not valid there.
• In implicit-differentiation problems, treating $y$ as $y(x)$ is part of the setup; then both sides must still be differentiable with respect to $x$.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: differentiate a one-point coincidence or a non-differentiable equation as though it were a differentiable relation → the new derivative equation is not justified and can be false.

Debug: ask “does this relation hold in the setting I am differentiating, and do both sides have derivatives there?” before writing a new line.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• Why does the move preserve equality only for a relation you are legitimately differentiating, rather than for every equation that happens to be true at one point?
• In an implicit equation like $x^2+y^2=25$, why does differentiating both sides create a $y'$ term instead of treating $y$ like a constant?

For the Principle

• How do you decide whether the equation is an identity, an implicit relation, or only an equation-to-solve? Why does that distinction decide whether differentiate-both-sides is even available?
• What changes if a relation holds on an interval but one side fails to be differentiable at an isolated point inside that interval?

Between Principles

• How does differentiate-both-sides differ from the derivative chain rule? One authorizes acting on the whole equation; the other authorizes differentiating a composite expression on one side. Where do they appear together in an implicit-differentiation chain?

Generate an Example

• Construct an equation that is true at one point but not as a relation in a neighborhood, and explain why differentiating both sides there would create a false derivative equation.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____If u(x)=v(x) is the relation you are differentiating, apply d/dx to both sides to get u'(x)=v'(x).

Write the canonical pattern: _____$u(x)=v(x) \Rightarrow u'(x)=v'(x)$

State the canonical condition: _____identity or implicit relation; u and v differentiable

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to use worked examples effectively.)

Procedure Walkthrough

Starting from $x^2 + y^2 = 25$, differentiate both sides with respect to $x$.

Assume $y$ is a differentiable function of $x$.

Step	Expression	Operation
0	$x^2 + y^2 = 25$	—
1	$\dfrac{d}{dx}(x^2 + y^2) = \dfrac{d}{dx}(25)$	Differentiate both sides — valid because $x^2+y^2=25$ is the relation being studied and both sides are differentiable with respect to $x$
2	$\dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(y^2) = 0$	Derivative sum rule on the left; derivative of a constant on the right
3	$2x + 2y\,y' = 0$	Power rule on $x^2$ and chain rule on $y^2$

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Drills

Format A: Forward step

Apply the move, then simplify the derivative expressions enough to make the new equation readable.

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Differentiate both sides of the identity

$\sin^2 x + \cos^2 x = 1$

Reveal

$\frac{d}{dx}(\sin^2 x+\cos^2 x)=\frac{d}{dx}(1) \Rightarrow 2\sin x\cos x - 2\cos x\sin x = 0$

This is legal because $\sin^2 x + \cos^2 x = 1$ is an identity, and each side is differentiable for all real $x$.

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Assume $y$ is differentiable as a function of $x$. Differentiate both sides of

$e^y = x + y$

Reveal

$\frac{d}{dx}(e^y)=\frac{d}{dx}(x+y) \Rightarrow e^y y' = 1 + y'$

This is a legal implicit-differentiation step: the relation is being studied as a curve, and both sides are differentiable with respect to $x$.

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Assume $y$ is differentiable as a function of $x$. Differentiate both sides of

$xy + y^2 = 6$

Reveal

$\frac{d}{dx}(xy+y^2)=\frac{d}{dx}(6) \Rightarrow x y' + y + 2y\,y' = 0$

Differentiate the whole relation first, then use the product rule and chain rule inside the new derivative equation.

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[Near-miss — negative] At $x=1$, can you differentiate both sides of

$x = x^2?$

Reveal

No. The displayed equation is true at $x=1$, but it is not the relation you are differentiating in any neighborhood of $x=1$.

If you tried the move anyway, you would get $1 = 2x$, and then $1=2$ at $x=1$. This is the key near-miss: one-point truth is not enough. Differentiate-both-sides acts on a differentiable relation, not on an isolated coincidence.

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Format B: Action label

Name the move being used, and say why it is or is not legal.

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What move takes this equation to the next line? Why is it valid?

$x^2 + y^2 = 16 \quad\longrightarrow\quad \frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(16)$

Reveal

Differentiate both sides. It is valid because $x^2+y^2=16$ is the relation being studied and both sides are differentiable with respect to $x$.

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What move is used first in the chain below?

$e^y = x + y \;\longrightarrow\; \frac{d}{dx}(e^y)=\frac{d}{dx}(x+y) \;\longrightarrow\; e^y y' = 1 + y'$

Reveal

The first move is differentiate both sides. After that, the later simplification uses the derivative sum rule on $x+y$ and the chain rule on $e^y$.

This distinction matters: the whole equation is transformed first; the term-by-term derivative rules come afterward.

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[Negative] A student writes

$x^3 + 1 = e^x \quad\longrightarrow\quad 3x^2 = e^x$

Is the move legal? Explain the exact condition failure.

Reveal

Illegal. This is an equation-to-solve, not an identity or implicit relation being differentiated.

Even though both expressions are differentiable, the move has no relation to act on here. Differentiability alone is not enough; the equality must be the differentiable relation under study.

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Format C: Transition identification and eligibility

Locate where the move appears, or decide whether the equation is eligible for it.

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Which transition uses differentiate both sides?

Transition	From	To
0→1	$x^2 + y^2 = 1$	$\dfrac{d}{dx}(x^2+y^2)=\dfrac{d}{dx}(1)$
1→2	$\dfrac{d}{dx}(x^2+y^2)=\dfrac{d}{dx}(1)$	$\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2)=0$
2→3	$\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2)=0$	$2x+2y\,y'=0$

Reveal

Transition 0→1 uses differentiate both sides.

Transition 1→2 uses the derivative sum rule and constant rule. Transition 2→3 uses the power rule and chain rule.

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[Negative] Which of these are eligible for differentiate-both-sides in the stated setting? Give a one-line reason for each.

1. $\sin^2 x + \cos^2 x = 1$ for all $x$
2. $x = x^2$ at $x=1$
3. $x^2 + y^2 = 9$, with $y$ differentiable as a function of $x$
4. $|x| = \sqrt{x^2}$ at $x=0$

Reveal

1. Eligible — it is an identity, and both sides are differentiable for all $x$.
2. Not eligible — it is true only at one isolated point, not as a relation in a neighborhood.
3. Eligible — it is the implicit relation being studied, and the setup assumes $y$ is differentiable.
4. Not eligible — although the equality is true, neither side is differentiable at $x=0$.

This is the real eligibility check: are you differentiating a genuine relation, and do both sides have derivatives in that setting?

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In the implicit-differentiation chain below, which step is the differentiate-both-sides step, and which step uses the derivative chain rule?

$x^3 + y^3 = 7 \,\longrightarrow\, \frac{d}{dx}(x^3+y^3)=\frac{d}{dx}(7) \,\longrightarrow\, 3x^2 + 3y^2 y' = 0$

Reveal

• The differentiate-both-sides step is the first arrow: $x^3 + y^3 = 7 \to \dfrac{d}{dx}(x^3+y^3)=\dfrac{d}{dx}(7)$.
• The chain rule appears in the second arrow when $\dfrac{d}{dx}(y^3)$ becomes $3y^2 y'$.

The whole-equation move and the within-expression move are different rules that appear back-to-back.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $x^2 + xy + y^2 = 12$, differentiate both sides with respect to $x$. Assume $y$ is differentiable as a function of $x$.

Full solution

Step	Expression	Move
0	$x^2 + xy + y^2 = 12$	—
1	$\dfrac{d}{dx}(x^2 + xy + y^2)=\dfrac{d}{dx}(12)$	Differentiate both sides
2	$\dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(xy) + \dfrac{d}{dx}(y^2) = 0$	Derivative sum rule and constant rule
3	$2x + (x y' + y) + 2y\,y' = 0$	Product rule on $xy$ and chain rule on $y^2$
4	$2x + y + (x + 2y)y' = 0$	Group the $y'$ terms as a genuine algebraic state change

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Related Principles

Principle	Relationship
Derivative at a point	Foundational meaning of a derivative; differentiate-both-sides assumes each side has a derivative to take
Derivative sum rule	Usually appears immediately after differentiating both sides when a side contains added terms
Derivative chain rule	Powers implicit differentiation: after differentiating both sides, composite terms such as $y^2$ and $e^y$ create the $y'$ factor through the chain rule

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FAQ

What does “differentiate both sides” mean?

It means applying $\dfrac{d}{dx}$ to the entire identity or implicit relation $u(x)=v(x)$ so that the next line is $u'(x)=v'(x)$. The move acts on the equation first; only after that do you simplify each derivative using rules like the sum rule, product rule, or chain rule.

When is it legal to differentiate both sides?

It is legal when the displayed equality is the differentiable relation you are studying and both sides are differentiable in that setting. If you are differentiating an identity, the identity must hold on the interval you are using. If you are differentiating an implicit relation, the relation is treated locally and each side must still have a derivative there.

Why is the move common in implicit differentiation?

Implicit equations like $x^2+y^2=25$ do not isolate $y$ first. Differentiating both sides turns the whole relation into a derivative equation, and then the chain rule converts derivatives of $y$-expressions into terms involving $y'$. That is why the move is often the first step in implicit differentiation.

What goes wrong when one side is not differentiable?

The transformed equation includes a derivative that does not exist, so the move has no theorem behind it. This is why near-miss examples like $|x|=\sqrt{x^2}$ at $x=0$ matter: the equality is true, but the condition fails, so you cannot differentiate both sides there.

Why can’t I use this on any equation that happens to be true at one point?

Because the move differentiates a relation, not a coincidence. An equation like $x=x^2$ is true at $x=1$, but it is not true as a relation in a neighborhood of $x=1$. Differentiating both sides would create the false new equation $1=2x$, which immediately breaks at that same point.

Do I simplify the equation before or after differentiating both sides?

Either can be correct, but they answer different tactical questions. If simplifying first reveals an easier identity, that can be better. If the goal is implicit differentiation or preserving a relation between variables, differentiating both sides first is often the cleanest entry point. The key is that the condition must still hold wherever you choose to apply the move.

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How This Fits in Unisium

Differentiate both sides is a move-selection principle about relation type as much as derivative mechanics. Unisium trains it as the entry step that converts an identity or implicit relation into a derivative equation, then layers follow-up rules such as the derivative sum rule and derivative chain rule to finish the chain. The goal is to make both guards automatic: do I have a real relation to differentiate, and do both sides have derivatives in this setting?

Explore further:

• Calculus Subdomain Map — Return to the calculus map to see where implicit differentiation begins inside the derivative application layer
• Derivative chain rule — the rule that creates $y'$ terms after differentiate-both-sides in implicit equations
• Self-Explanation — use the drills above to explain not only what move is legal, but why tempting alternatives fail
• Retrieval Practice — make the equation and condition retrievable on demand

Ready to master differentiate both sides? Start practicing with Unisium or explore the full learning framework in Masterful Learning.

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