Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: implicit differentiation (201-SN2-RE)

Implicit differentiation calls for no new rule: it is the chain rule, applied to a letter you had got used to treating as a constant. That is where the whole difficulty lies, and the marks lost in this chapter are counted in missing $y'$ factors.

This sheet is written for CEGEP students in the Science program (Sciences de la nature) taking Differential Calculus, 201-SN2-RE (formerly 201-NYA). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

In the equation of a curve, $y$ is not a letter, it is a FUNCTION of $x$ : every time you meet it, the chain rule leaves a $y'$ behind. The whole chapter comes down to that factor, then to reading the quotient you obtain.

This chapter is part of 201-SN2-RE, Differential Calculus

The essentials

The principle: $y$ is a function of $x$ , not a letter

• Differentiate BOTH sides of the equation with respect to $x$ , treating $y$ as $y(x)$ .
• Every term containing $y$ therefore produces a factor $y'$ by the chain rule, exactly like $u'$ in $\left(u^{3}\right)'=3u^{2}u'$ .
• Then gather all the $y'$ terms on one side, factor, divide: the result is a QUOTIENT in $x$ and $y$ .
• An equation such as $x^{2}+y^{2}=25$ defines no function on $[-5, 5]$ , but it does define one AROUND every point where $y\neq 0$ . That is what makes the computation legitimate.

The line

x=3

meets the circle at TWO points: the equation does not define a function, but it defines one around each point, and the two slopes

-\frac{3}{4}

and

+\frac{3}{4}

are opposite.

Checking count before going on: as many $y'$ in the differentiated line as there are terms containing $y$ in the original equation.

The five structures that leave a $y'$

• $\frac{d}{dx}\left(y^{n}\right)=ny^{n-1}y'$ , and likewise $\frac{d}{dx}\left(\sqrt{y}\right)=\frac{y'}{2\sqrt{y}}$ .
• $\frac{d}{dx}(\sin y)=\cos y\cdot y'$ , $\frac{d}{dx}\left(e^{y}\right)=e^{y}y'$ , $\frac{d}{dx}(\ln y)=\frac{y'}{y}$ .
• PRODUCT: $\frac{d}{dx}(xy)=y+xy'$ . Both terms, always.
• QUOTIENT: $\frac{d}{dx}\left(\frac{x}{y}\right)=\frac{y-xy'}{y^{2}}$ .
• A term that contains only $x$ is differentiated as usual, and a constant gives $0$ : it is the only case with no $y'$ .

Reading the quotient: three kinds of zeros, three verdicts

• $y'$ generally depends on $x$ AND $y$ , and that is normal: the slope depends on the POINT, not on the $x$ -value alone.
• NUMERATOR alone zero: HORIZONTAL tangent.
• DENOMINATOR alone zero: VERTICAL tangent. The slope does not exist, but the tangent does, and its equation is $x=x_{0}$ .
• BOTH zero together: SINGULAR point, a crossing or a cusp. There may be two tangents, or none.

A point found by one of these conditions must still BELONG to the curve: always substitute into the original equation to obtain it.

Second derivative and conics: the two formulas that save time

• $y''$ is computed by differentiating $y'$ like any other expression, WITHOUT forgetting that $y'$ still depends on $x$ .
• Substitute the expression of $y'$ AFTER differentiating, then simplify with the equation of the curve: on $x^{2}+y^{2}=r^{2}$ this gives $y''=-\frac{r^{2}}{y^{3}}$ .
• Tangent to a conic in point form: for $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at the point $(x_{0}, y_{0})$ , it is $\frac{xx_{0}}{a^{2}}+\frac{yy_{0}}{b^{2}}=1$ . The minus sign of the hyperbola is kept.
• This form remains valid at points with a vertical tangent, where the quotient $y'$ says nothing any more.

\frac{x^{2}}{9}+\frac{y^{2}}{4}=1

: at

\left(\frac{9}{5}, \frac{8}{5}\right)

the formula gives

x+2y=5

, and at

(3, 0)

the vertical line

x=3

, where

y'

does not exist.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Differentiating $y$ as if it were an inert letter

the whole question

What not to write

“ $\frac{d}{dx}\left(y^{3}\right)=3y^{2}$ ”

What to write

“ $\frac{d}{dx}\left(y^{3}\right)=3y^{2}y'$ , because $y$ is a function of $x$ and the chain rule applies.”

Why: This is the mistake that defines the chapter. A paper that makes it once makes it in every term, and the result no longer has anything to do with the curve.

2. Forgetting the product rule on a mixed term

2 marks, and a wrong quotient all the way to the end

What not to write

“ $\frac{d}{dx}(xy)=xy'$ ”

What to write

“ $\frac{d}{dx}(xy)=y+xy'$ : it is a product of two functions of $x$ , so both terms.”

Why: The mixed term is the one read the fastest. The remedy is to circle it in the question before differentiating: wherever $x$ and $y$ touch, there will be two terms.

3. Solving for $y$ before differentiating, and losing a branch

half the marks of the question

What not to write

“ $x^{2}+y^{2}=25$ gives $y=\sqrt{25-x^{2}}$ , so $y'=\frac{-x}{\sqrt{25-x^{2}}}$ and the slope at $x=3$ is $-\frac{3}{4}$ .”

What to write

“ $y'=-\frac{x}{y}$ : at $(3, 4)$ the slope is $-\frac{3}{4}$ , but at $(3, -4)$ it is $+\frac{3}{4}$ . The square root kept only half of the circle.”

y=\sqrt{25-x^{2}}

describes only the upper half circle (green). The point

(3, -4)

, with its slope

+\frac{3}{4}

, is on the lost half (red);

y'=-\frac{x}{y}

covers both.

Why: Solving for $y$ picks a branch without saying so. Implicit differentiation handles both at once, and that is exactly why it exists.

4. Believing that a $y$ left in the result is the sign of a mistake

wasted time, and often a correct answer erased

What not to write

“My result still contains $y$ , I must have forgotten to finish.”

What to write

“ $y'=-\frac{x}{y}$ depends on $x$ and $y$ : it is the normal form of an implicit derivative, since the slope depends on the POINT and not on the $x$ -value alone.”

Why: An implicit curve passes twice through the same $x$ -value with two different slopes: no expression in $x$ alone could describe both.

5. Announcing a vertical tangent without looking at the numerator

2 marks, and a wrong geometric conclusion

What not to write

“The denominator of $y'$ is zero at the origin, so the tangent there is vertical.”

What to write

“Numerator AND denominator are zero at the origin: the point is SINGULAR. The curve crosses itself there and has two tangents, neither of which is given by the quotient.”

On the folium

x^{3}+y^{3}=6xy

, the numerator alone is zero at the point with a horizontal tangent, the denominator alone at the point with a vertical tangent, and BOTH at the origin, where the curve crosses itself.

Why: The quotient $\frac{0}{0}$ decides nothing, here as anywhere else. Look at both zeros before concluding anything about the tangent.

6. Computing $y''$ while forgetting that $y'$ still depends on $x$

the whole concavity question

What not to write

“ $y'=-\frac{x}{y}$ , so $y''=-\frac{1}{y}$ .”

What to write

“ $y''=-\frac{y-xy'}{y^{2}}$ , then replace $y'$ by $-\frac{x}{y}$ : $y''=-\frac{y^{2}+x^{2}}{y^{3}}=-\frac{25}{y^{3}}$ on the circle of radius $5$ .”

Why: The quotient is differentiated with the quotient rule, and the $y$ in the denominator produces yet another $y'$ . Substituting before differentiating makes that term disappear.

7. Concluding that there is no tangent when the slope does not exist

1 mark, and often the next question

What not to write

“At $(5, 0)$ , $y'=-\frac{5}{0}$ does not exist, so the circle has no tangent at this point.”

What to write

“The SLOPE does not exist, the TANGENT does: it is the vertical line $x=5$ .”

(5, 0)

y'=-\frac{5}{0}

does not exist, but the circle does have a tangent: the vertical line

x=5

. No slope, but a tangent.

Why: A vertical tangent line is perfectly well defined but has no slope. Confusing the two amounts to confusing the geometric object with its equation $y=mx+b$ .

8. Redoing the computation curve by curve in a family

the time of the exam, and the expected general proof

What not to write

“I differentiate $x^{2}+y^{2}=9$ , then $x^{2}+y^{2}=16$ , then $x^{2}+y^{2}=25$ ...”

What to write

“ $x^{2}+y^{2}=c$ gives $y'=-\frac{x}{y}$ for the WHOLE family at once: the constant disappears when you differentiate.”

y'=-\frac{x}{y}

holds for the whole family

x^{2}+y^{2}=c

: on the same radius, the tangents to the three circles are parallel, with slope

-\frac{3}{4}

. The constant

c

disappeared when differentiating.

Why: This is the real strength of the method: the slope is expressed in terms of the POINT and not of the chosen curve, which lets you prove that two families are orthogonal in one line.

Which method to choose

What to do with the quotient, depending on what the question asks

Once $y'=\frac{N(x,y)}{D(x,y)}$ is obtained, the verb of the question tells you what to do with it. Never start the computation over.

If “the slope at the point $(x_{0}, y_{0})$ ” → first check that the point is ON the curve, then substitute both coordinates

Example: on $x^{2}+xy+y^{2}=7$ , the point $(1, 2)$ gives $y'=-\frac{4}{5}$
If “the points where the tangent is horizontal” → solve $N=0$ TOGETHER WITH the equation of the curve, then check $D\neq 0$

Example: $2x+y=0$ with $x^{2}+xy+y^{2}=7$ gives $3x^{2}=7$

Solving $N=0$ alone gives a whole line, not points: the system with the equation of the curve is compulsory.
If “the points where the tangent is vertical” → same system with $D=0$ , then check $N\neq 0$ ; the tangent is written $x=x_{0}$

Example: $x+2y=0$ with the same conic gives $3y^{2}=7$
If $N$ and $D$ vanish at the same point → singular point: conclude nothing from the quotient, describe the curve another way (parametrization, symmetry, factoring the equation)

Example: the origin of the folium $x^{3}+y^{3}=6xy$
If “the equation of the tangent line” → the slope, then $y-y_{0}=m(x-x_{0})$ ; on a conic in standard form, the point form $\frac{xx_{0}}{a^{2}}+\frac{yy_{0}}{b^{2}}=1$ is faster

Example: $4x+5y=14$ for the conic above at the point $(1, 2)$
If “the concavity” or “ $y''$ ” → differentiate the quotient $y'$ with the quotient rule, then replace $y'$ by its expression and simplify with the equation of the curve

If the question asks for a property that holds for a whole FAMILY of curves, do not fix the constant: it disappears when you differentiate, and that is what lets you conclude in one go.

How to differentiate each structure of the equation

Sweep through the equation term by term and sort each term into one of these lines, before writing anything.

If the term contains only $x$ , or it is a constant → ordinary differentiation, no $y'$

Example: $\frac{d}{dx}\left(3x^{2}-7\right)=6x$
If the term is a power or a root of $y$ → power rule, then $\times\,y'$

Example: $\frac{d}{dx}\left(y^{4}\right)=4y^{3}y'$
If the term is a transcendental function of $y$ → the line of the table, then $\times\,y'$

Example: $\frac{d}{dx}\left(\sin y\right)=\cos y\cdot y'$
If $x$ and $y$ touch in the same term → product or quotient rule, which gives TWO terms

Example: $\frac{d}{dx}\left(x^{2}y^{3}\right)=2xy^{3}+3x^{2}y^{2}y'$

This is the line students skip. Spotting these terms in the question before starting costs five seconds and saves two marks.

Once every term is differentiated, only algebra is left: gather the $y'$ terms, factor, divide. No subtlety of differential calculus beyond that line.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

The equation of the tangent line to an implicit curve

When to use it: The question gives the equation of a curve and a point, and asks for the tangent line or the normal line.

1 Check that the point belongs to the curve, and SAY so: “ $1^{2}+1\times 2+2^{2}=7$ , the point $(1, 2)$ is indeed on the curve.”
2 Differentiate both sides with respect to $x$ , writing $y'$ for every term containing $y$ : “ $2x+y+xy'+2yy'=0$ .”
3 Gather and factor: “ $y'(x+2y)=-(2x+y)$ , so $y'=-\frac{2x+y}{x+2y}$ .”
4 Substitute both coordinates, never the $x$ -value alone: “ $y'(1, 2)=-\frac{4}{5}$ .”
5 Write the equation of the tangent line and simplify it: “ $y-2=-\frac{4}{5}(x-1)$ , that is $4x+5y=14$ .”

Concluding sentence

The tangent line to the curve $x^{2}+xy+y^{2}=7$ at the point $(1, 2)$ has equation $4x+5y=14$ .

The trap: Substituting $x=1$ as early as the second line. You then lose the general formula for $y'$ , which the next question almost always asks for, and the computation becomes impossible to check.

Marking: Usually 1 mark for the complete differentiation with its $y'$ factors, 1 mark for solving for $y'$ , 1 mark for the numerical slope, 1 mark for the final equation.

Proving that two families of curves are orthogonal

When to use it: The question gives two families depending on a parameter and asks you to show that they cross at right angles.

1 Differentiate the first family implicitly WITHOUT fixing the parameter: “ $x^{2}+y^{2}=c$ gives $2x+2yy'=0$ , so $m_{1}=-\frac{x}{y}$ .”
2 Do the same for the second: “ $y=kx$ gives $k=\frac{y}{x}$ , so $m_{2}=k=\frac{y}{x}$ .”
3 Note that both slopes depend only on the common POINT, and not on the parameters $c$ and $k$ .
4 Compute the product: “ $m_{1}m_{2}=-\frac{x}{y}\times\frac{y}{x}=-1$ for every point with $x\neq 0$ and $y\neq 0$ .”
5 Conclude by naming the criterion used, and point out the excluded points.

Concluding sentence

At every point of intersection other than the origin, the product of the slopes is $-1$ : the two families cross at right angles, whatever the values of the parameters.

The trap: Eliminating the parameter BEFORE differentiating, by replacing $k$ with its numerical value. The proof then holds for one curve only, and the question asked for infinitely many.

Marking: 1 mark for each slope obtained by implicit differentiation, 1 mark for the product and the conclusion, often 1 mark for mentioning the excluded points.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Check that the point is on the curve, before anything else

Substitute both coordinates into the ORIGINAL equation and compare the two sides. Three seconds, and it immediately exposes a question copied incorrectly.

On $x^{2}+xy+y^{2}=7$ , the point $(1, 2)$ gives $1+2+4=7$ : it is fine. The point $(2, 1)$ would also give $7$ , but $(1, 3)$ would give $13$ , and everything that follows would have to be thrown away.
Count the $y'$ factors before factoring

Count the terms that contain $y$ in the original equation, then count the terms carrying a $y'$ in the differentiated line. The two numbers must match.

$x^{2}+xy+y^{2}=7$ contains two terms in $y$ , so the differentiated line must carry two $y'$ : $2x+y+xy'+2yy'=0$ . A single $y'$ signals a skipped product rule.
Check the sign of the slope on a sketch

Place the point roughly on the curve and see whether it rises or falls. The sign found must agree, and this check takes less than ten seconds.

On the circle, the point $(3, 4)$ is at the top right, and the curve falls there: a positive slope would be wrong. At $(3, -4)$ , at the bottom right, it rises.
Confirm the slope with a nearby point of the curve

Choose a nearby $x$ , solve the equation of the curve for the $y$ on the SAME branch, then compute the average rate of change between the two points.

On the circle of radius $5$ , between $(3, 4)$ and $(3.01, 3.99249)$ , the rate is about $-0.751$ , which confirms $-\frac{3}{4}$ .

The typical problem, taken apart

A conic: tangent line at a point, then horizontal tangents

Consider the curve with equation $x^{2}+xy+y^{2}=7$ .

1. Give the equation of the tangent line at the point $(1, 2)$ .

2. Find the points of the curve where the tangent is horizontal.

The curve

x^{2}+xy+y^{2}=7

is the graph of no function, but around

(1, 2)

it defines one: that is where implicit differentiation goes to find the slope.

Step 1

$1^{2}+1\times 2+2^{2}=1+2+4=7$ : the point $(1, 2)$ does belong to the curve.

Why

This line is often worth half a mark and above all it protects you: a tangent computed at a point off the curve is wrong whatever the rest of the computation.

Step 2

Differentiate both sides with respect to $x$ : $2x+\left(y+xy'\right)+2yy'=0$ .

Why

The term $xy$ is a PRODUCT of two functions of $x$ , so it gives two terms. The term $y^{2}$ gives $2yy'$ , never $2y$ . Two terms in $y$ on the left, two $y'$ on the right: the count is right.

Step 3

Gather: $y'(x+2y)=-(2x+y)$ , so $y'=-\dfrac{2x+y}{x+2y}$ .

Why

From here on there is no more differential calculus, only algebra. This general expression will serve both questions, and keeping it saves redoing everything.

Step 4

At $(1, 2)$ : $y'=-\dfrac{2+2}{1+4}=-\dfrac{4}{5}$ , hence $y-2=-\dfrac{4}{5}(x-1)$ , that is $4x+5y=14$ .

Why

Substitute BOTH coordinates. Substituting the $x$ -value alone would leave a $y$ in the slope, and the other point of the curve with $x$ -coordinate $1$ has a different slope.

Step 5

Horizontal tangent: $2x+y=0$ , that is $y=-2x$ . Substituting into the equation: $x^{2}-2x^{2}+4x^{2}=3x^{2}=7$ , so $x=\pm\sqrt{\dfrac{7}{3}}$ .

Why

Solve $N=0$ AS A SYSTEM with the equation of the curve. Solving $2x+y=0$ on its own would give a whole line, almost none of whose points are on the curve.

Step 6

The two points are $\left(\sqrt{\dfrac{7}{3}}, -2\sqrt{\dfrac{7}{3}}\right)\approx(1.53, -3.06)$ and its symmetric point $(-1.53, 3.06)$ . At each of them, $x+2y=-3\sqrt{\dfrac{7}{3}}\neq 0$ .

Why

The check $D\neq 0$ is compulsory: if the denominator vanished at the same place, the point would be singular and the conclusion “horizontal tangent” would be wrong.

Step 7

Check at $(1.53, -3.06)$ : $(1.53)^{2}+(1.53)(-3.06)+(-3.06)^{2}\approx 2.34-4.68+9.36=7.02\approx 7$ , and likewise at the symmetric point.

Why

Substituting the points found back into the original equation is the most rewarding check of the chapter: it catches a sign error as well as a badly extracted root.

The conclusion, written out

The tangent line at the point $(1, 2)$ has equation $4x+5y=14$ , and the tangent is horizontal at the two points $\left(\pm\sqrt{\frac{7}{3}}, \mp 2\sqrt{\frac{7}{3}}\right)$ , about $(1.53, -3.06)$ and $(-1.53, 3.06)$ .

The classic mistake on this problem: Differentiating $xy$ as $xy'$ only. You then get $y'=-\frac{2x}{x+2y}$ , so $-\frac{2}{5}$ at the point $(1, 2)$ instead of $-\frac{4}{5}$ : the tangent is twice too flat, and question 2 gives points that are not on the curve.

Learn by heart

• As many $y'$ in the differentiated line as there are terms containing $y$ in the equation.
• $\frac{d}{dx}(xy)=y+xy'$ : the mixed term ALWAYS gives two terms.
• $y'$ depends on $x$ AND $y$ , and that is the normal form of the result.
• Numerator alone zero: horizontal tangent. Denominator alone zero: vertical tangent, with equation $x=x_{0}$ .
• Both zero together: singular point, the quotient says nothing more.
• A candidate point is found AS A SYSTEM with the equation of the curve, never on its own.
• For $y''$ , differentiate the quotient, then replace $y'$ : on $x^{2}+y^{2}=r^{2}$ , $y''=-\frac{r^{2}}{y^{3}}$ .
• A family is differentiated without fixing its parameter: the constant disappears and the slope depends only on the point.

Frequently asked questions

Why do you add y prime when you differentiate y?

Because y is not a constant: the equation defines y as a function of x, even if you cannot write it down. Differentiating y squared is therefore the same as differentiating u squared with u a function of x, which gives two u times u prime by the chain rule. The y prime is exactly that u prime.

How do you find the points where the tangent line is horizontal?

Set the numerator of y prime equal to zero, but solve that condition as a system with the equation of the curve: otherwise you get a whole line, not points. Then check that the denominator is not zero at the same place, otherwise the point would be singular rather than a point with a horizontal tangent.

Is it normal for the derivative to still contain y?

Yes, it is even the rule. An implicit curve often passes twice through the same x-value, with two different slopes: no expression in x alone could describe both. The slope depends on the whole point, so on x and y, and you substitute both coordinates at the end.

What do you do when the denominator of y prime is zero?

Look at the numerator at the same point. If it is not zero, the tangent is vertical and its equation is x equals the x-coordinate of the point: the slope does not exist but the tangent does. If the numerator is zero as well, the point is singular, the curve crosses itself or has a cusp there, and the quotient decides nothing any more.

Can you solve for y before differentiating?

When it is possible, yes, but you then pick a single branch without noticing it. On a circle, writing y equals the square root of twenty-five minus x squared keeps only the upper half, and every slope of the lower half comes out with the wrong sign. Implicit differentiation handles both branches at once.

Practise it

Corrected exercises: Implicit differentiation, 201-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

16 corrected exercises
160 points
225 minutes

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Revision sheet: implicit differentiation (201-SN2-RE)

The essentials

The principle: yyy is a function of xxx, not a letter

The five structures that leave a y′y'y′

Reading the quotient: three kinds of zeros, three verdicts

Second derivative and conics: the two formulas that save time

The mistakes that cost marks

1. Differentiating yyy as if it were an inert letter

2. Forgetting the product rule on a mixed term

3. Solving for yyy before differentiating, and losing a branch

4. Believing that a yyy left in the result is the sign of a mistake

5. Announcing a vertical tangent without looking at the numerator

6. Computing y′′y''y′′ while forgetting that y′y'y′ still depends on xxx

7. Concluding that there is no tangent when the slope does not exist

8. Redoing the computation curve by curve in a family

Which method to choose

What to do with the quotient, depending on what the question asks

How to differentiate each structure of the equation

How the answer is expected to be written

The equation of the tangent line to an implicit curve

Proving that two families of curves are orthogonal

Check before you hand in

The typical problem, taken apart

A conic: tangent line at a point, then horizontal tangents

Learn by heart

Frequently asked questions

Why do you add y prime when you differentiate y?

How do you find the points where the tangent line is horizontal?

Is it normal for the derivative to still contain y?

What do you do when the denominator of y prime is zero?

Can you solve for y before differentiating?

Corrected exercises: Implicit differentiation, 201-SN2-RE

See also

Looking for a 201-SN2-RE tutor in Montreal?

The principle: $y$ is a function of $x$ , not a letter

The five structures that leave a $y'$

1. Differentiating $y$ as if it were an inert letter

3. Solving for $y$ before differentiating, and losing a branch

4. Believing that a $y$ left in the result is the sign of a mistake

6. Computing $y''$ while forgetting that $y'$ still depends on $x$