Differential Calculus
Implicit Differentiation
Most curves worth studying are not graphs of functions: a circle, an ellipse, the folium . Implicit differentiation computes slopes on such curves without ever solving for — you differentiate the whole equation with respect to while treating as an unknown differentiable function of . It is nothing more than the chain rule applied inside an equation, and it is the machinery behind related rates, logarithmic differentiation, and the derivatives of inverse functions later in the course.
Explicit versus implicit
A function is given explicitly when it is written , with already isolated. An equation in and — write it — is said to define implicitly as a function of near a point if there is an open interval containing and a function on with
One equation can define many functions. The circle contains both (upper semicircle) and (lower semicircle); near only the first one applies, near only the second. Near neither works: no interval around carries a function on the curve at all. That failure is not an accident, and the next theorem says exactly when it can happen.
Why implicit differentiation works
Theorem (Implicit Function Theorem, one-variable form). Let have partial derivatives and that exist and are continuous on an open disc around . Suppose
- (the point is on the curve), and
- .
Then there are an open interval containing and a unique differentiable function on with and on , and
(Here means "differentiate treating as a constant" and means "differentiate treating as a constant".)
Where the formula comes from. Granted that such a differentiable exists, the function is identically on , so . The chain rule for a function of two variables gives , hence and whenever .
Two consequences you must respect:
- Implicit differentiation assumes differentiability. Hypothesis 2 is what supplies it. Where the method can produce a meaningless expression (division by zero) or the curve can genuinely fail to be a function of — a vertical tangent, a cusp, or a self-crossing.
- The answer normally contains both and . That is correct and expected: the slope depends on which branch of the curve you are standing on.
The mechanics
Every time a is differentiated, the chain rule attaches a factor (written below).
| Term appearing in the equation | Its derivative with respect to |
|---|---|
| constant | |
| (needs ) | |
| (needs ) | |
| (product rule) | |
| (quotient rule) | |
The four-step recipe. (1) Differentiate both sides with respect to , applying the chain rule to every . (2) Move every term containing to one side and everything else to the other. (3) Factor out . (4) Divide, and record which points make the divisor zero — those are exactly the candidates for vertical tangents and singular points.
Worked example 1 — circle: tangent line and second derivative
Curve at the point , which is on the curve since .
At : , so the tangent line is
Independent check. Near the branch is , whose derivative is — the same formula. Also, the distance from the origin to the line is , the radius, as a tangent to a circle must be.
Now the second derivative. Differentiate again with the quotient rule, remembering is a function of :
using the original equation in the last step. At , : the upper semicircle is concave down, as it should be.
Worked example 2 — folium of Descartes
Curve . The point lies on it: .
At : , so the tangent is , i.e. .
Horizontal tangent. Need and and the point on the curve. From , substitute into :
So or , giving . At the denominator is , so there is a genuine horizontal tangent there. Because swapping and leaves the equation unchanged, the vertical tangent sits at the mirror point .
The origin. At both numerator and denominator vanish, and : the Implicit Function Theorem does not apply. The curve genuinely crosses itself there (a node) with two tangent lines, and . No single value of exists.
Horizontal and vertical tangents
Write the implicit derivative as a quotient and test only points that satisfy the original equation:
| Situation at a point on the curve | Conclusion |
|---|---|
| , | horizontal tangent |
| , | vertical tangent () |
| , | inconclusive: cusp, node, or isolated point — investigate separately |
The step students skip is substituting back into . Solving alone gives a curve of candidates, most of which are nowhere near the graph.
Second derivative implicitly
Differentiate the expression for with respect to (product/quotient rule, chain rule on every ), then substitute the known and finally use the original equation to simplify. Example: for , and
Common mistakes
| Wrong | Right | Why |
|---|---|---|
| is a function of ; the chain rule is not optional. | ||
| or | is a product of two functions of . | |
| Same omission, most often on circles. | ||
| Differentiating only the left side | Differentiate both sides | An equation stays an equation only if both sides get the same operator; , not . |
| Leaving scattered on both sides | Collect, factor, then divide | must become |
| Reporting as "wrong because it has in it" | It is the correct form | Implicit derivatives legitimately depend on both coordinates. |
| Setting the numerator to and reporting those points | Also substitute into | Candidates must lie on the curve; the folium's horizontal tangent is at , not at every point with . |
| The denominator contains , so the quotient rule applies. | ||
| Using at on | The tangent is vertical, | : the theorem's hypothesis fails and no exists. |
| Differentiating and reporting | The equation has no real solutions | The method presupposes a real curve with a differentiable branch; formal symbol pushing proves nothing. |
Key terms
- implicitly defined function
- explicit form
- chain rule
- product rule
- Implicit Function Theorem
- partial derivative
- branch of a curve
- tangent line
- horizontal tangent
- vertical tangent
- singular point (cusp, node)
- second derivative
- folium of Descartes
- lemniscate
- astroid
- rotated ellipse
Practice Problems
For the circle , find by implicit differentiation, then give the slope of the curve and the equation of the tangent line at the point .
Show hint
Differentiate both sides with respect to ; the term needs the chain rule, so a factor of appears.
Show answer
First confirm the point is on the curve: . Good.
Differentiate both sides with respect to , treating as a function of :
At :
Tangent line through with slope :
Check. Near the curve is the lower semicircle , whose derivative is ; at this is . Also the distance from the origin to is , the radius — exactly what a tangent line to this circle must satisfy.
Answer: ; slope ; tangent line .
The equation defines implicitly as a function of for . (a) Find by implicit differentiation. (b) Solve for explicitly, differentiate, and confirm the two answers agree.
Show hint
On the left-hand side you have a product of two functions of , so the product rule comes before anything else.
Show answer
(a) Implicitly. Differentiate both sides with respect to . The left side is a product of and , so use the product rule :
(b) Explicitly. From with we get , so by the power rule
Agreement. Substitute into the implicit answer:
The two expressions are identical. This is the whole point of the method: the implicit answer is the same derivative, just written in terms of both coordinates.
Numerical spot check at : implicit gives ; explicit gives . A centred difference on with step gives .
Answer: ; the two methods agree.
Find for the curve , and evaluate it at the point .
Show hint
Differentiate term by term; the term picks up a from the chain rule, and the right-hand side is a constant.
Show answer
Check the point first: , so is on the curve.
Differentiate both sides with respect to :
At :
Independent check. Here can be isolated: . By the chain rule,
At this is , matching.
Note also that everywhere the formula is defined, so the curve is decreasing — sensible, since and must trade off to keep the sum equal to .
Answer: , which equals at .
The curve passes through . Find , give the tangent line at , and state where on the curve the formula is valid.
Show hint
Write the square roots as powers, and , before differentiating; then think about which points make a denominator zero.
Show answer
Check the point: . Good.
Write and differentiate with respect to :
At :
so the tangent line is , that is
Validity. The curve lives in (since ) with . The derivative formula requires (division by ) and the differentiation of requires , i.e. . So is valid exactly on . At the endpoint the tangent is vertical; at it is horizontal, and neither is covered by the formula.
Independent check. Explicitly, , so . At this is , matching. In general , so the two forms agree everywhere on .
Answer: ; tangent at is ; valid for .
Find for the curve and compute the slope of the tangent line at .
Show hint
Both terms are products of a power of with a power of , so each one needs the product rule and the chain rule.
Show answer
Check the point: . Good.
Differentiate each term with respect to . For use the product rule with , :
For use , (and the chain rule on ):
Adding, and setting the result equal to :
At :
Numerical check. Solving for near : at the quadratic gives ; at the quadratic gives . The centred slope is , agreeing with to the accuracy of the step size.
Answer: ; the slope at is .
Find the equation of the tangent line to the ellipse at the point .
Show hint
The constants and just ride along as denominators; differentiate as usual and then isolate .
Show answer
Check the point: . Good.
Differentiate both sides with respect to :
At :
Tangent line:
Independent check. The standard tangent formula for at is . Here that reads , i.e. . Solving for : — the same line. Substituting gives , so the line does pass through the point.
Answer: , with slope .
The curve passes through . Find and the tangent line there. Explain why this curve has no vertical tangent lines.
Show hint
Differentiate the left side with the chain rule: the inside function is , whose derivative is .
Show answer
Check the point: and . Good.
Differentiate both sides with respect to . On the left, the chain rule with inner function gives an inner derivative :
At we have , so
and the tangent line is
No vertical tangents. A vertical tangent needs the denominator to vanish. But , so : the denominator is never . Hence is defined at every point of the curve, and in fact it is trapped between and , so the curve is strictly increasing with slope in — no horizontal tangents either.
Numerical check. Take and solve for near . Writing and , the equation becomes . With : , so , giving . (Solving numerically gives .) Then , confirming the slope.
Answer: ; at the slope is and the tangent is ; no vertical tangents because .
For the curve , show that .
Show hint
Find first, then differentiate that quotient again — and remember to substitute the expression you already found for before simplifying.
Show answer
Step 1: first derivative. Differentiate with respect to :
Step 2: differentiate again. Apply the quotient rule to , treating as a function of (so ):
Step 3: substitute .
Step 4: use the original equation. Since ,
and therefore
which is what we wanted.
Numerical check at . Then , so and , giving . Directly from we get , , , so the second difference is , consistent.
Answer: and .
The rotated ellipse is a closed curve. Find every point on it where the tangent line is horizontal and every point where the tangent line is vertical.
Show hint
Get as a single fraction, then handle numerator and denominator separately — and in each case remember the point must also satisfy the original equation.
Show answer
Step 1: the implicit derivative. Differentiate , using the product rule on :
Step 2: horizontal tangents need numerator and denominator . From we get . Substituting into the curve:
This gives and . Check the denominators: at , ; at , . Both are genuine horizontal tangents, namely the lines and .
Step 3: vertical tangents need denominator and numerator . From we get . Substituting:
This gives and . Check the numerators: at , ; at , . Both are genuine vertical tangents, the lines and .
Equivalently, at these points , which is the precise statement that the tangent is vertical.
Verification that all four points are on the curve. : . : . : . : . All correct. Note that the numerator and denominator never vanish simultaneously on the curve (that would force , which does not satisfy ), so there are no singular points.
Answer: horizontal tangents at and ; vertical tangents at and .
Find the equation of the tangent line to the lemniscate at the point .
Show hint
The left side is a composite: differentiate the outer square first, then the inner . Substituting the numbers , early keeps the algebra short.
Show answer
Step 0: the point is on the curve. and , so the left side is and the right side is . Good.
Step 1: differentiate. On the left, the chain rule on gives outer derivative times inner derivative :
Step 2: substitute , where :
Step 3: the tangent line.
that is .
General formula (useful as a check). Expanding step 1 and collecting:
At : .
Numerical check. Setting and , the curve becomes . At () the relevant root gives ; at () it gives . The centred slope is , matching .
Answer: slope ; tangent line .
Find the equation of the tangent line to the astroid at the point . Then identify the four points where the astroid fails to have an ordinary tangent slope, and say what happens there.
Show hint
Use the power rule on and (the second needs the chain rule). It helps to note that , so its cube root is exactly .
Show answer
Step 0: the point is on the curve. and , so the sum is . Good.
Step 1: differentiate.
Step 2: evaluate. Since , its cube root is . And . Hence
Step 3: the tangent line.
Check: at , the line gives . Correct.
Numerical check of the slope. On the branch near : at , ; at , . The centred slope is , matching .
Step 4: the bad points. The differentiation step divides by and the resulting formula divides by , so the work breaks down where or . (Note , so the intercepts are at .)
- If then , so : the points and . There is undefined (division by ) while , so the tangent line is vertical.
- If then , so : the points and . Near the curve consists of the two branches , both defined only for , so is not a function of on any interval around and no two-sided derivative exists. The upper branch approaches with slope and the lower branch with the opposite sign, ; both tend to as , so the tangent line there is horizontal, .
All four points are cusps. Parametrize the astroid as , (then ); the velocity vanishes exactly at , i.e. exactly at these four points, and the curve reverses direction there instead of passing smoothly through. The Implicit Function Theorem is silent at all four: for the partials and blow up at and respectively, so the continuity hypothesis fails.
Answer: ; the tangent at is ; the four cusps are , where the tangent line is horizontal but no two-sided exists, and , where the tangent is vertical.
Let and let be a point on the circle . (a) Assuming , prove that the tangent line at is . (b) What happens when , and why does the implicit method not deliver an answer there? (c) Formally differentiating also produces . Explain why this computation is meaningless.
Show hint
For (a) run the usual implicit differentiation with general letters and clear denominators at the end, using the fact that satisfies the equation. For (b) and (c), look back at the hypotheses of the Implicit Function Theorem.
Show answer
(a) Differentiate with respect to (note is a constant):
At with the slope is , so the tangent line is
Since lies on the circle, , so the tangent line is
Check. The distance from the origin to this line is , so the line touches the circle exactly once — the defining property of a tangent to a circle. Also the line's direction vector is perpendicular to the radius vector , since , which is the classical geometric fact.
(b) If then , so the point is or — the far left or far right of the circle. The tangent there is the vertical line , which has no slope. The implicit formula correctly refuses to produce a number: it involves division by .
The deeper reason is the Implicit Function Theorem. With we have , so and hypothesis 2 fails. And indeed the conclusion fails too: no interval around carries any function whose graph lies on the circle, because for there are no points of the circle at all, while for just below there are two. The curve is locally a function of , not of : writing and differentiating gives at , which is the correct statement that the tangent is vertical.
Note the tangent formula from part (a) still gives the right answer by continuity: with it reads , i.e. . So the formula is valid at all points of the circle, even though the derivation required .
(c) The equation has no real solutions: for all real . There is no curve, no point , and hence no function with to differentiate.
Implicit differentiation is not a rule for transforming equations; it is the chain rule applied to an identity that is assumed to hold for a real differentiable on an interval. When no such exists, every step of the manipulation is vacuous — the symbols can be pushed around, but they refer to nothing. This is precisely why the Implicit Function Theorem opens with the hypothesis at a real point: you must know a solution exists before you may differentiate it.
Answer: (a) . (b) At the tangent is vertical () and , so the theorem's hypothesis and its conclusion both fail. (c) The equation has no real solutions, so there is no differentiable branch to differentiate and the computation is empty.