MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: implicit differentiation (MATH 203)

Q: How do you solve for dy/dx after implicit differentiation?

Move every term that contains dy/dx to one side of the equation and every other term to the other side, changing the sign of each term that crosses the equal sign. Then factor dy/dx out of all its terms and divide by what is left. If a dy/dx sits inside a fraction, first multiply every term by that denominator.

Q: When should I plug in the point in implicit differentiation?

If the question asks only for the slope or the tangent at one given point, substitute the coordinates right after differentiating: the equation in dy/dx becomes a numerical equation, much easier to solve. For the second derivative, differentiate twice first and substitute only at the end, otherwise terms disappear.

Q: How do I find where an implicit curve has a horizontal tangent?

Write dy/dx as one fraction. Set the numerator equal to zero, which gives a line or a curve, then substitute it into the original equation to find the actual points. Keep only the points where the denominator is not zero. For vertical tangents, do the same with the denominator and check the numerator.

Q: Why does my calculator give an error for a negative number to the power 2/3?

Many scientific calculators compute fractional powers through logarithms, which need a positive base. The cube root of a negative number does exist: take the cube root first, then square it. For example, minus 8 to the power 2/3 is the square of minus 2, which is 4.

This sheet is not a summary of section 3.7 of Thomas' Calculus: you already have the course notes. It answers one question only, what makes students lose marks on implicit differentiation in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The calculus of this chapter fits in one line: differentiate both sides, and every $y$ -term leaves a factor $y'$ . The marks go in the next lines, which are algebra: collecting, factoring, clearing fractions, rewriting exponents, substituting the equation of the curve. That is where this sheet spends its time. A scientific calculator is allowed on the exam, but it neither differentiates nor simplifies, and every value below has the exact form the marker expects.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

Implicit differentiation is one line of calculus followed by a linear equation in $y'$ : the marks are lost in the algebra that follows, a sign lost when a term crosses the equal sign, a $y'$ left on the other side, a fraction or a negative exponent mishandled, a parameter or a curve equation not used at the point.

This chapter is part of MATH 203, Calculus I (Concordia)

The essentials

One line of calculus, then a linear equation in y'

• Step 1 (calculus): differentiate BOTH sides with respect to $x$ , $y$ being a function of $x$ . Every term containing $y$ leaves a factor $y'$ : $\frac{d}{dx}(y^n) = ny^{n-1}y'$ , $\frac{d}{dx}(xy) = y + xy'$ , $\frac{d}{dx}\frac{1}{y} = -\frac{y'}{y^2}$ , $\frac{d}{dx}\sqrt y = \frac{y'}{2\sqrt y}$ .
• Step 2 (algebra): collect ALL the $y'$ terms on one side, changing the sign of every term that crosses the equal sign; factor $y'$ out; divide.
• If only the slope at one point is asked, substitute the coordinates right after step 1: the equation in $y'$ then has numbers only.
• Horizontal tangent: numerator $N = 0$ and denominator $D \ne 0$ ; vertical: $D = 0$ and $N \ne 0$ . The condition is a line or a curve; the answer is where it meets the given curve.
• Normal line at a point: slope $-\frac{1}{m}$ when the tangent has slope $m \ne 0$ .

x^2 + 2xy + 2y^2 = 4

, the red line

N = 0

(

y = -x

) meets the curve at the two horizontal-tangent points, the green line

D = 0

(

x = -2y

) at the two vertical-tangent points.

Writing 'differentiate both sides with respect to $x$ ' and checking that the given point satisfies the equation are worth method marks on their own, and the check takes five seconds.

The algebra that comes with the chapter

• A negative exponent moves ITS factor to the other floor: $\frac{x^{-2/3}}{y^{-2/3}} = \frac{y^{2/3}}{x^{2/3}}$ .
• A fraction inside the equation is cleared by multiplying EVERY term by its denominator; a complex fraction is cleared by multiplying its numerator AND its denominator by the same quantity.
• $x^{2/3}$ for $x < 0$ is $\left(\sqrt[3]{x}\right)^2 > 0$ : take the cube root first.
• At a point of the curve, the coordinates satisfy its equation: use it to simplify $y''$ , or to eliminate a parameter $a$ from a slope.

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. Losing the sign when a term crosses the equal sign

2 to 3 marks, and a tangent that goes the wrong way

What not to write

“ $4x + y + xy' + 2yy' = 0$ , so $xy' + 2yy' = 4x + y$ and $y' = \frac{6}{5}$ at $(1, 2)$ .”

What to write

“ $xy' + 2yy' = -4x - y$ , so $y' = -\frac{4x + y}{x + 2y}$ , which is $-\frac{6}{5}$ at $(1, 2)$ .”

Why: The calculus was right; the collecting step moved $4x + y$ without changing its sign. On the curve $2x^2 + xy + y^2 = 8$ the arc through $(1, 2)$ goes down, so a positive slope is impossible.

2. Leaving a y' on the other side

the whole question

What not to write

“ $2x + 2yy' = y' + 3$ , so $y' = \frac{3 - 2x}{2y}$ .”

What to write

“ $2yy' - y' = 3 - 2x$ , so $y'(2y - 1) = 3 - 2x$ and $y' = \frac{3 - 2x}{2y - 1}$ .”

Why: Every $y'$ term must be on the same side before factoring. At $x = 1$ , $y = 2$ the student gets $\frac{1}{4}$ instead of $\frac{1}{3}$ .

3. Turning a negative-exponent fraction upside down

2 marks, and a slope sixteen times too steep

What not to write

“ $y' = -\frac{x^{-2/3}}{y^{-2/3}} = -\left(\frac{x}{y}\right)^{2/3}$ , so the slope of $\sqrt[3]{x} + \sqrt[3]{y} = 3$ at $(8, 1)$ is $-4$ .”

What to write

“ $-\frac{x^{-2/3}}{y^{-2/3}} = -\frac{y^{2/3}}{x^{2/3}} = -\left(\frac{y}{x}\right)^{2/3}$ , which is $-\frac{1}{4}$ at $(8, 1)$ .”

(8, 1)

the curve

\sqrt[3]{x} + \sqrt[3]{y} = 3

is almost flat: the true tangent (green, slope

-\frac{1}{4}

) follows it, the wrong one (red, slope

-4

) cuts straight through.

Why: A negative exponent sends its own factor across the fraction bar: $x^{-2/3}$ goes down, $y^{-2/3}$ comes up. The wrong answer is the reciprocal of the right one.

4. Giving a line as the answer to a horizontal-tangent question

3 marks

What not to write

“On $x^2 + 2xy + 2y^2 = 4$ , $y' = -\frac{x + y}{x + 2y} = 0$ when $y = -x$ : the tangent is horizontal along $y = -x$ .”

What to write

“ $y = -x$ and $x^2 + 2xy + 2y^2 = 4$ give $x^2 = 4$ : the points $(2, -2)$ and $(-2, 2)$ , where $x + 2y = \mp 2 \ne 0$ .”

Why: The condition $N = 0$ describes a line of the plane; the curve crosses it at two points only. The figure of the essentials shows the line and its two points.

5. Clearing the complex fraction of y'' on one floor only

2 marks

What not to write

“On $3x^2 - y^2 = 11$ , $y'' = \frac{3y - \frac{9x^2}{y}}{y^2} = \frac{3y^2 - 9x^2}{y^2}$ .”

What to write

“Multiplying numerator AND denominator by $y$ : $y'' = \frac{3y^2 - 9x^2}{y^3} = -\frac{33}{y^3}$ on the curve.”

Why: Multiplying one floor of a fraction changes its value. At $(2, 1)$ the error is invisible because $y = 1$ ; at $(3, 4)$ it gives $-\frac{33}{16}$ instead of $-\frac{33}{64}$ .

6. Leaving a parameter in a slope compared at a point

the conclusion of the question

What not to write

“The slopes of $x^2 + y^2 = ax$ and $x^2 + y^2 = by$ are $\frac{a - 2x}{2y}$ and $\frac{2x}{b - 2y}$ ; their product depends on $a$ and $b$ , so the circles are not always orthogonal.”

What to write

“At a common point, $a = \frac{x^2 + y^2}{x}$ and $b = \frac{x^2 + y^2}{y}$ , so the slopes are $\frac{y^2 - x^2}{2xy}$ and $\frac{2xy}{x^2 - y^2}$ , of product $-1$ .”

Why: At a point of a curve, its parameter is fixed by the point. A slope written with the parameter still in it hides the equation of the curve, which is part of the data.

7. Reading the derivative of 1/y as 1/y'

the whole term, and every line after it

What not to write

“ $\frac{d}{dx}\frac{1}{y} = \frac{1}{y'}$ .”

What to write

“ $\frac{1}{y} = y^{-1}$ , so $\frac{d}{dx}\frac{1}{y} = -y^{-2}y' = -\frac{y'}{y^2}$ .”

Why: The derivative of a reciprocal is not the reciprocal of the derivative. With $y = x^2$ at $x = 1$ : the true value is $-2$ , the student's is $\frac{1}{2}$ .

8. Letting the calculator decide a cube root

the whole question

What not to write

“The slope at $(-1, 64)$ is $-\frac{64^{2/3}}{(-1)^{2/3}}$ , and the calculator gives an error: the slope does not exist.”

What to write

“ $(-1)^{2/3} = \left(\sqrt[3]{-1}\right)^2 = 1$ and $64^{2/3} = 16$ , so the slope is $-16$ .”

Why: Many scientific calculators compute powers through logarithms and refuse a negative base. The cube root of a negative number exists: take it first, then square.

Which method to choose

Which algebraic move, by the shape of the line in front of you

Look at the equation right after differentiating, before touching it

If $y'$ terms on BOTH sides of the equal sign → bring every $y'$ term to the left, change the sign of what crosses, factor $y'$ out

Example: $2x + 2yy' = 2y'$ gives $y'(2y - 2) = -2x$ , so $y' = -\frac{x}{y - 1}$
If a $y'$ inside a fraction, such as $\frac{xy'}{2\sqrt y}$ → multiply EVERY term by the denominator before solving

Example: $x\sqrt y + y = 10$ : $2y + xy' + 2\sqrt y\,y' = 0$ , so $y' = -\frac{2y}{x + 2\sqrt y}$ , which is $-\frac{8}{7}$ at $(3, 4)$
If fractions in $x$ and $y$ in the equation itself → clear the denominators BEFORE differentiating, then factor if possible

Example: $\frac{x}{y} + \frac{y}{x} = \frac{5}{2}$ becomes $(2x - y)(x - 2y) = 0$ : two lines, slopes $2$ and $\frac{1}{2}$
If negative or fractional exponents in the result → move each negative power to the other floor, one factor at a time

Example: $-\frac{x^{-2/3}}{y^{-2/3}} = -\left(\frac{y}{x}\right)^{2/3}$ , which is $-\frac{1}{4}$ at $(8, 1)$
If only the slope at ONE given point → check the point, substitute right after differentiating, solve a numerical equation

Example: $x^2y^2 + 3y = 2x + 8$ at $(1, 2)$ : $8 + 7y' = 2$ , so $y' = -\frac{6}{7}$
If $y''$ asked → quotient rule with $y$ a function, replace $y'$ , clear the complex fraction on both floors, use the equation of the curve

Example: $3x^2 - y^2 = 11$ : $y'' = \frac{3y^2 - 9x^2}{y^3} = -\frac{33}{y^3}$
If a parameter $a$ in the equation of a family → replace $a$ by its value at the point, read from the equation

Example: $x^2 + y^2 = ax$ : $a = \frac{x^2 + y^2}{x}$ and the slope is $\frac{y^2 - x^2}{2xy}$

If the formula for y' reads 0 over 0 at the point, no branch applies: the point is a crossing, a cusp or something else, and the curve must be studied near it.

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.

Finding the points with a horizontal or vertical tangent

When to use it: Any question that says 'find all the points where the tangent is horizontal' or 'vertical'

1 Differentiate both sides with respect to $x$ and solve for $y'$ in the form $\frac{N}{D}$ , simplified by any common factor.
2 Horizontal: set $N = 0$ and solve it for one variable; this is a line or a curve, NOT the answer.
3 Substitute into the equation of the curve and solve: these are the candidate points.
4 Check $D \ne 0$ at each candidate, and say so; a candidate where $D = 0$ too gives $\frac{0}{0}$ and is excluded.
5 Vertical: the same four steps with $D = 0$ , checking $N \ne 0$ .

Concluding sentence

“The tangent is horizontal where $x + y = 0$ and $x + 2y \ne 0$ . Substituting $y = -x$ into $x^2 + 2xy + 2y^2 = 4$ gives $x^2 = 4$ , so the points are $(2, -2)$ and $(-2, 2)$ , where $x + 2y = \mp 2 \ne 0$ : the tangent is horizontal at these two points.”

The trap: Stopping at 'the tangent is horizontal when $y = -x$ ', or forgetting to check the other part of the fraction.

Marking: Typically 1 mark for $y'$ , 1 for the condition, 2 for the substitution and the points, 1 for the check of the other part.

Check before you hand in

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

The point is on the curve

Substitute the given point into the equation before differentiating. A point that is not on the curve makes every slope meaningless, and some equations have no points at all.

$(1, 2)$ on $2x^2 + xy + y^2 = 8$ : $2 + 2 + 4 = 8$ . The equation $x^2 + y^2 + 1 = 0$ has no real point, so no slope to compute.
The sign agrees with the picture

Compare the sign of the slope with the direction of the curve at the point, on the figure or on a quick sketch. A sign lost in the collecting step shows here and nowhere else.

At $(1, 2)$ the curve $2x^2 + xy + y^2 = 8$ goes down: $y' = -\frac{6}{5}$ is plausible, $+\frac{6}{5}$ is not.
Test a formula at a point where y is not 1

A forgotten power of y is invisible at a point with y = 1 or with x = 0. Test a formula for y' or y'' at a second, generic point of the curve.

On $3x^2 - y^2 = 11$ , $\frac{3y^2 - 9x^2}{y^2}$ agrees with $-\frac{33}{y^3}$ at $(2, 1)$ but gives $-\frac{33}{16}$ instead of $-\frac{33}{64}$ at $(3, 4)$ .
Perpendicular slopes multiply to -1

For a normal line or for orthogonal curves, multiply the two slopes before moving on: the product must be -1.

Tangent $-\frac{6}{5}$ and normal $\frac{5}{6}$ at $(1, 2)$ : product $-1$ . A normal of slope $-\frac{5}{6}$ would give $1$ .

The typical problem, taken apart

Tangent, normal and second derivative at one point of a cubic curve

The curve $x^2y + xy^2 = 6$ passes through $P(1, 2)$ . Find the tangent and the normal lines at $P$ , and the value of $y''$ at $P$ .

Exact answers, every step justified as on a MATH 203 final.

The branch of

x^2y + xy^2 = 6

in the first quadrant, and the point

P(1, 2)

on it: the tangent and the normal there are what the question asks for.

Step 1

Check: $1 \cdot 2 + 1 \cdot 4 = 6$ , so $P$ is on the curve. Differentiate both sides with respect to $x$ , product rule on each term, chain rule on $y^2$ : $2xy + x^2y' + y^2 + 2xyy' = 0$ .

Why

Both terms are products of two functions of $x$ ; naming the rules shows the marker that each $y'$ was placed on purpose.

Step 2

Substitute $x = 1$ , $y = 2$ before isolating: $4 + y' + 4 + 4y' = 0$ , so $5y' = -8$ and $y' = -\frac{8}{5}$ .

Why

Only the slope at $P$ is asked: the general formula $y' = -\frac{2xy + y^2}{x^2 + 2xy}$ is not needed, and the collecting step becomes arithmetic.

Step 3

Tangent: $y - 2 = -\frac{8}{5}(x - 1)$ , that is $y = -\frac{8}{5}x + \frac{18}{5}$ . Normal, slope $\frac{5}{8}$ : $y - 2 = \frac{5}{8}(x - 1)$ , that is $y = \frac{5}{8}x + \frac{11}{8}$ .

Why

The normal slope is the NEGATIVE reciprocal; the check $-\frac{8}{5} \cdot \frac{5}{8} = -1$ costs one line.

Step 4

Differentiate $2xy + x^2y' + y^2 + 2xyy' = 0$ once more, $y$ and $y'$ being functions of $x$ : $2y + 2xy' + 2xy' + x^2y'' + 2yy' + 2yy' + 2x(y')^2 + 2xyy'' = 0$ , that is $2y + 4xy' + 4yy' + 2x(y')^2 + (x^2 + 2xy)y'' = 0$ .

Why

Differentiating the differentiated equation avoids a quotient rule and a complex fraction; the product rule on $2xy \cdot y'$ gives three terms, the usual place to lose one.

Step 5

At $P$ with $y' = -\frac{8}{5}$ : $4 - \frac{32}{5} - \frac{64}{5} + \frac{128}{25} + 5y'' = 0$ . Over $25$ : $\frac{100 - 160 - 320 + 128}{25} = -\frac{252}{25}$ , so $5y'' = \frac{252}{25}$ and $y'' = \frac{252}{125}$ .

Why

The numbers go in only AFTER the second differentiation, and a common denominator keeps the arithmetic exact: $\frac{252}{125} = 2.016$ .

Step 6

Verification with the explicit branch $y = \frac{-x^2 + \sqrt{x^4 + 24x}}{2x}$ (the curve is quadratic in $y$ ): at $x = 1$ , $y = \frac{-1 + 5}{2} = 2$ , and a table of values at $x = 0.99$ and $1.01$ gives a slope close to $-1.6$ .

Why

A second route, even a numerical one on the calculator, catches a sign lost in the collecting step.

The conclusion, written out

“ $P(1, 2)$ is on the curve. Implicit differentiation gives $y' = -\frac{8}{5}$ at $P$ , so the tangent is $y = -\frac{8}{5}x + \frac{18}{5}$ and the normal is $y = \frac{5}{8}x + \frac{11}{8}$ . Differentiating again and substituting, $y'' = \frac{252}{125}$ at $P$ .”

The classic mistake on this problem: Writing $\frac{d}{dx}(xy^2) = y^2 + 2yy'$ , without the factor $x$ on the second term: at $P$ this gives $4 + y' + 4 + 4y' = 0$ by luck, since $x = 1$ , and the error only appears in $y''$ .

Learn by heart

• Differentiate BOTH sides with respect to $x$ ; every $y$ -term leaves a factor $y'$ .
• Collect ALL the $y'$ terms on one side, changing the sign of what crosses, factor $y'$ , then divide.
• $\frac{d}{dx}(xy) = y + xy'$ and $\frac{d}{dx}\frac{1}{y} = -\frac{y'}{y^2}$ , never $\frac{1}{y'}$ .
• Horizontal: $N = 0$ , $D \ne 0$ ; vertical: $D = 0$ , $N \ne 0$ ; always at points OF THE CURVE.
• $y''$ : differentiate $y'$ with $y$ still a function, clear the complex fraction on BOTH floors, use the equation.
• $\frac{a^{-n}}{b^{-n}} = \frac{b^n}{a^n}$ , and $x^{2/3} = \left(\sqrt[3]{x}\right)^2$ even for $x < 0$ .

Frequently asked questions

How do you solve for dy/dx after implicit differentiation?

Move every term that contains dy/dx to one side of the equation and every other term to the other side, changing the sign of each term that crosses the equal sign. Then factor dy/dx out of all its terms and divide by what is left. If a dy/dx sits inside a fraction, first multiply every term by that denominator.

When should I plug in the point in implicit differentiation?

If the question asks only for the slope or the tangent at one given point, substitute the coordinates right after differentiating: the equation in dy/dx becomes a numerical equation, much easier to solve. For the second derivative, differentiate twice first and substitute only at the end, otherwise terms disappear.

How do I find where an implicit curve has a horizontal tangent?

Write dy/dx as one fraction. Set the numerator equal to zero, which gives a line or a curve, then substitute it into the original equation to find the actual points. Keep only the points where the denominator is not zero. For vertical tangents, do the same with the denominator and check the numerator.

Why does my calculator give an error for a negative number to the power 2/3?

Many scientific calculators compute fractional powers through logarithms, which need a positive base. The cube root of a negative number does exist: take the cube root first, then square it. For example, minus 8 to the power 2/3 is the square of minus 2, which is 4.

Practise it

Corrected exercises: Implicit differentiation, MATH 203 at Concordia

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.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet The chain rule Next sheet Derivatives of inverse functions and logarithms

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-implicit-differentiation. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: implicit differentiation (MATH 203)

The essentials

One line of calculus, then a linear equation in y'

The algebra that comes with the chapter

The mistakes that cost marks

1. Losing the sign when a term crosses the equal sign

2. Leaving a y' on the other side

3. Turning a negative-exponent fraction upside down

4. Giving a line as the answer to a horizontal-tangent question

5. Clearing the complex fraction of y'' on one floor only

6. Leaving a parameter in a slope compared at a point

7. Reading the derivative of 1/y as 1/y'

8. Letting the calculator decide a cube root

Which method to choose

Which algebraic move, by the shape of the line in front of you

How the answer is expected to be written

Finding the points with a horizontal or vertical tangent

Check before you hand in

The typical problem, taken apart

Tangent, normal and second derivative at one point of a cubic curve

Learn by heart

Frequently asked questions

How do you solve for dy/dx after implicit differentiation?

When should I plug in the point in implicit differentiation?

How do I find where an implicit curve has a horizontal tangent?

Why does my calculator give an error for a negative number to the power 2/3?

Corrected exercises: Implicit differentiation, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?