MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: implicit differentiation and inverse trigonometric derivatives (MATH 140)

Q: How do you do implicit differentiation step by step?

Differentiate both sides of the equation with respect to x, treating y as a function of x, so that every term containing y gets a factor dy/dx from the chain rule and every mixed term like xy needs the product rule. Then move all the dy/dx terms to one side, factor dy/dx out, and divide. If only the slope at one point is asked, substitute the coordinates before isolating.

Q: Why does my dy/dx answer still contain y?

Because an implicit curve can have several points above the same value of x, each with its own slope. The circle x squared plus y squared equals 10 has the points (1, 3) and (1, -3), with slopes -1/3 and 1/3. The y in the formula tells the points apart, so to get a number you substitute both coordinates of a point of the curve.

Q: How do I find horizontal and vertical tangent lines of an implicit curve?

Write dy/dx as a fraction. For horizontal tangents, set the numerator to zero, solve that condition together with the equation of the curve, and keep the points where the denominator is not zero. For vertical tangents, do the same with the denominator, keeping the points where the numerator is not zero. If both vanish, the formula gives no conclusion.

Q: How do you prove the derivative of arcsin x with implicit differentiation?

Write y = arcsin x as sin y = x with y between -pi/2 and pi/2. Differentiate: cos y times dy/dx = 1, so dy/dx = 1 / cos y. Since cos y is not negative on that interval, cos y equals the square root of 1 - x squared, and the derivative is 1 over the square root of 1 - x squared, for x strictly between -1 and 1.

Q: How do I find the derivative of an inverse function at a point?

Find the number a such that f(a) = b, usually by inspection or from a table, then compute 1 divided by f prime of a. Do not evaluate f prime at b: b is an output of f, not an input. The formula needs f prime of a to be nonzero; where it is zero, the inverse has a vertical tangent.

This sheet is not a summary of section 3.5 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on implicit differentiation and inverse trigonometric derivatives in MATH 140 at McGill University, and which precise gesture avoids each loss.

The computations of this chapter are short, and that is the danger: a single missing $y'$ or a slope evaluated at the wrong point costs the whole question, and a proof of $\frac{d}{dx}\arcsin x$ without its sign argument costs half of it. Every value below is exact and done by hand, as on the exam.

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

$y$ is a function of $x$ that you cannot see: every $y$ -term differentiated leaves a factor $y'$ , and every answer belongs to a POINT of the curve, never to an $x$ alone. For inverse functions the same rule reads: the range of $\arcsin$ fixes the sign of the square root, and $(f^{-1})'(b)$ is computed at the point $a$ where $f(a) = b$ .

This chapter is part of MATH 140, Calculus 1 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Implicit differentiation
Implicitly defined curve
Tangent and normal lines
Horizontal and vertical tangents
Implicit second derivative
Derivative of arcsin
Derivative of arccos
Derivative of arctan
Derivative of an inverse function
Orthogonal curves
Chain rule

The essentials

y is a function you cannot see: every y-term leaves a y'

• Implicit differentiation: differentiate BOTH sides of the equation with respect to $x$ , treating $y$ as a function $y(x)$ . Every term that contains $y$ is a composition with inner function $y$ , and the chain rule leaves the factor $y' = \frac{dy}{dx}$ .
• $\frac{d}{dx}(y^n) = ny^{n-1}y'$ , $\frac{d}{dx}(\sin y) = \cos y \cdot y'$ , $\frac{d}{dx}(e^y) = e^yy'$ , and by the product rule $\frac{d}{dx}(xy) = y + xy'$ .
• The result $y' = \frac{N(x, y)}{D(x, y)}$ belongs to a POINT, not to an $x$ : on $x^2 + y^2 = 10$ , the value $x = 1$ carries the points $(1, 3)$ and $(1, -3)$ , and two different slopes.
• Horizontal tangent: $N = 0$ and $D \ne 0$ at a point OF THE CURVE. Vertical tangent: $D = 0$ and $N \ne 0$ . When both vanish, the formula decides nothing.
• $y''$ : differentiate $y'$ again with $y$ still a function of $x$ , substitute $y'$ , then simplify with the equation of the curve.

One vertical line

x = 1

, two points of the circle, two tangents: slope

-\frac{1}{3}

(1, 3)

and

\frac{1}{3}

(1, -3)

. That is why

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

must contain

y

The first line of the copy, 'differentiate both sides with respect to $x$ ', and the check that the given point satisfies the equation, are worth method marks on their own.

Inverse functions: the range fixes the sign, the right point fixes the value

• $y = \arcsin x$ means $\sin y = x$ with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ ; $y = \arccos x$ means $\cos y = x$ with $0 \le y \le \pi$ ; $y = \arctan x$ means $\tan y = x$ with $-\frac{\pi}{2} < y < \frac{\pi}{2}$ .
• Differentiating these equations: $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ and $\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$ ; $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$ for every $x$ .
• The sign of each square root comes from the RANGE: $\cos y \ge 0$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , $\sin y \ge 0$ on $[0, \pi]$ .
• Inverse function: if $f(a) = b$ and $f'(a) \ne 0$ , then $(f^{-1})'(b) = \frac{1}{f'(a)}$ . The derivative of $f$ is taken at $a = f^{-1}(b)$ , never at $b$ .
• $\sin^{-1}x$ means $\arcsin x$ , never $\frac{1}{\sin x}$ , which is $\csc x$ .

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Differentiating each kind of term with respect to x

Read a line as: the term of the first column, differentiated by the rule of the second, gives the third. The red lines are derivatives that students write and that do not exist: each one removes the unknown $y'$ from a place where it belongs.

Term	Rule	Derivative
$y^3$	chain, inner $y$	$3y^2y'$
Example: $x^2 + y^3 = 9$ at $(1, 2)$ : $2 + 12y' = 0$ , so $y' = -\frac{1}{6}$ .
$xy$	product	$y + xy'$
Example: $xy = 6$ at $(2, 3)$ : $3 + 2y' = 0$ , so $y' = -\frac{3}{2}$ , as $y = \frac{6}{x}$ confirms.
$\sin y$	chain, inner $y$	$\cos y \cdot y'$
Example: $x + \sin y = 1$ at $(1, 0)$ : $1 + \cos 0 \cdot y' = 0$ , so $y' = -1$ .
$e^{xy}$	chain, then product	$e^{xy}(y + xy')$
Example: At $x = 0$ , $y = 5$ , $y' = 1$ : $e^0(5 + 0 \cdot 1) = 5$ .
$y^3$	as if $y$ were $x$	$3y^2$ no such rule
Example: On $x^2 + y^3 = 9$ at $(1, 2)$ it gives $2 + 12 = 0$ : an equation with no $y'$ to solve for. What to do: Multiply by $y'$ : the derivative of $g(y)$ is $g'(y)\,y'$ .
$x^2y$	derivative of each factor	$2xy'$ no such rule
Example: On $x^2y + y^3 = 10$ at $(3, 1)$ it forces $y' = 0$ , while the true slope is $-\frac{1}{2}$ . What to do: Product rule: $\frac{d}{dx}(x^2y) = 2xy + x^2y'$ .

Test for any term: freeze $x$ and ask whether it contains $y$ . If yes, its derivative contains $y'$ exactly once per occurrence of $y$ .

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. Forgetting the factor y' on a y-term

the whole question: without y' there is nothing to isolate

What not to write

“Differentiating $x^2 + y^3 = 9$ : $2x + 3y^2 = 0$ .”

What to write

“ $2x + 3y^2y' = 0$ , so $y' = -\frac{2x}{3y^2}$ , and at $(1, 2)$ , $y' = -\frac{1}{6}$ .”

Why: $y^3$ is a composition with inner function $y(x)$ . The student's line is the equation of another curve, not a derivative.

2. Skipping the product rule on a mixed term

2 to 3 marks, and an answer that says the curve is flat

What not to write

“ $\frac{d}{dx}(x^2y) = 2xy'$ , so on $x^2y + y^3 = 10$ : $y'(2x + 3y^2) = 0$ and $y' = 0$ .”

What to write

“ $2xy + x^2y' + 3y^2y' = 0$ , so $y' = -\frac{2xy}{x^2 + 3y^2}$ , which is $-\frac{1}{2}$ at $(3, 1)$ .”

Why: $x^2y$ is a product of two functions of $x$ . The points $(1, 2)$ and $(3, 1)$ are both on the curve, so it cannot be flat: a slope $0$ everywhere is the signal to recheck each product.

3. Giving a slope for an x instead of a point

half the question

What not to write

“On $x^2 + y^2 = 10$ , the slope at $x = 1$ is $-\frac{1}{3}$ .”

What to write

“At $x = 1$ the circle has two points, $(1, 3)$ and $(1, -3)$ , where $y' = -\frac{x}{y}$ is $-\frac{1}{3}$ and $\frac{1}{3}$ .”

Why: An implicit curve can have several points above one $x$ , and the slope formula contains $y$ precisely to tell them apart. The figure of the essentials shows both tangents.

4. Answering a horizontal-tangent question with a line

3 marks

What not to write

“On $x^2 - xy + y^2 = 3$ , $y' = \frac{y - 2x}{2y - x} = 0$ when $y = 2x$ : the tangents are horizontal along $y = 2x$ .”

What to write

“ $y = 2x$ and $x^2 - xy + y^2 = 3$ give $3x^2 = 3$ : the points $(1, 2)$ and $(-1, -2)$ , where $2y - x = \pm 3 \ne 0$ .”

Why: $N = 0$ describes a line, and the curve crosses it at a few points only. The answer is the intersection, and each point needs the check $D \ne 0$ .

5. Reading a zero denominator as a vertical tangent

2 marks

What not to write

“On $y^2 = x^2 - x^4$ , $y' = \frac{x(1 - 2x^2)}{y}$ has a zero denominator at the origin, so the tangent there is vertical.”

What to write

“At the origin both $N$ and $D$ vanish: no conclusion. Near $0$ the curve is $y = \pm x\sqrt{1 - x^2}$ , two branches of slopes $1$ and $-1$ . At $(1, 0)$ , $N = -1 \ne 0$ : that tangent is vertical.”

The same zero denominator, two stories: at the origin two branches cross with slopes

1

and

-1

, at

(1, 0)

the tangent is vertical because the numerator is

-1

Why: $D = 0$ gives a vertical tangent only when $N \ne 0$ . A $\frac{0}{0}$ usually means that several branches pass through the point, each with its own slope.

6. Differentiating y' as if y were a constant

3 marks

What not to write

“On $4x^2 + 9y^2 = 36$ , $y' = -\frac{4x}{9y}$ , so $y'' = -\frac{4}{9y}$ .”

What to write

“ $y'' = -\frac{4}{9} \cdot \frac{y - xy'}{y^2} = -\frac{4(4x^2 + 9y^2)}{81y^3} = -\frac{16}{9y^3}$ .”

Why: $y$ is still a function of $x$ in the second derivative. At $\left(\frac{3}{2}, \sqrt 3\right)$ the wrong answer is only $\frac{3}{4}$ of the right one; at $(0, 2)$ the two agree, so test at a point with $x \ne 0$ .

7. Evaluating the inverse derivative at the wrong point

the whole question

What not to write

“ $f(x) = x^3 + x$ , so $(f^{-1})'(2) = \frac{1}{f'(2)} = \frac{1}{13}$ .”

What to write

“ $f(1) = 2$ , so $f^{-1}(2) = 1$ and $(f^{-1})'(2) = \frac{1}{f'(1)} = \frac{1}{4}$ .”

The tangent to

f

(1, 2)

has slope

4

; its reflection in

y = x

, the tangent to

f^{-1}

(2, 1)

, has slope

\frac{1}{4}

, not

\frac{1}{13}

Why: $b = 2$ is a VALUE of $f$ . The slope of $f$ must be read at the input $a$ that produces it, and the tangent to $f^{-1}$ at $(2, 1)$ is the reflection of the tangent to $f$ at $(1, 2)$ .

8. Leaving the sign of the square root to chance

2 marks, half of the proof

What not to write

“ $\cos y = \pm\sqrt{1 - x^2}$ , and we keep the plus sign, so $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ .”

What to write

“Since $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ , $\cos y \ge 0$ , so $\cos y = \sqrt{1 - x^2}$ and $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ .”

Why: The question asks for a PROOF, and the range of $\arcsin$ is the only reason for the sign. For $\arccos$ the range is $[0, \pi]$ , and the argument moves to $\sin y \ge 0$ .

9. Reading sin to the power minus 1 as a reciprocal

the whole question

What not to write

“ $\sin^{-1}x = \frac{1}{\sin x}$ , so $\frac{d}{dx}\sin^{-1}x = -\frac{\cos x}{\sin^2 x}$ .”

What to write

“ $\sin^{-1}x = \arcsin x$ , so $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$ , which is $1$ at $x = 0$ .”

Why: A $-1$ written on the NAME of a function means the inverse function; the reciprocal is written $(\sin x)^{-1}$ or $\csc x$ . The student's formula is not even defined at $x = 0$ .

Which method to choose

Which move, by what the question asks

Read the verb and the object of the question before differentiating anything

If the slope or the tangent line at a GIVEN point → check the point is on the curve, differentiate, substitute the coordinates, then solve the linear equation in y'

Example: $2(x^2 + y^2)^2 = 25(x^2 - y^2)$ at $(3, 1)$ : $240 + 80y' = 150 - 50y'$ , so $y' = -\frac{9}{13}$
If the points with a HORIZONTAL tangent → set the numerator of $y'$ to $0$ , intersect with the curve, check the denominator

Example: $x^2 - xy + y^2 = 3$ : $y = 2x$ gives $(1, 2)$ and $(-1, -2)$
If the points with a VERTICAL tangent → set the denominator to $0$ , intersect with the curve, check the numerator

Example: $x^2 - xy + y^2 = 3$ : $x = 2y$ gives $(2, 1)$ and $(-2, -1)$
If $y''$ , or $\frac{d^2y}{dx^2}$ → differentiate $y'$ with $y$ a function, substitute $y'$ , simplify with the equation

Example: $4x^2 + 9y^2 = 36$ : $y'' = -\frac{16}{9y^3}$
If $(f^{-1})'(b)$ with no formula for $f^{-1}$ → find $a$ with $f(a) = b$ by inspection, then $\frac{1}{f'(a)}$

Example: $f(x) = x^3 + 2x + 1$ , $b = 4$ : $a = 1$ , answer $\frac{1}{5}$
If the derivative of $\arcsin$ , $\arccos$ or $\arctan$ of an inner function → the formula at the inner function, times the derivative of the inner function

Example: $\frac{d}{dx}\arctan(2x) = \frac{2}{1 + 4x^2}$
If two curves meet at right angles → find the common points, both slopes there, and show the product is $-1$

Example: $x^2 + y^2 = 9$ and $(x - 5)^2 + y^2 = 16$ at $\left(\frac{9}{5}, \frac{12}{5}\right)$ : $-\frac{3}{4} \cdot \frac{4}{3} = -1$

If the formula for $y'$ gives $\frac{0}{0}$ at the point, no branch applies: study the curve near the point, often by solving for $y$ locally.

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.

Proving the derivative of an inverse trigonometric function

When to use it: Any question that says 'prove' or 'derive' the formula for $\frac{d}{dx}\arcsin x$ , $\frac{d}{dx}\arccos x$ or $\frac{d}{dx}\arctan x$

1 Rewrite $y = \arcsin x$ as $\sin y = x$ AND state the range $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ : the range is part of the definition.
2 Differentiate both sides with respect to $x$ , naming the chain rule: $\cos y \cdot y' = 1$ .
3 Solve for $y'$ : $y' = \frac{1}{\cos y}$ , where $\cos y \ne 0$ .
4 Write $\cos y$ in terms of $x$ with $\cos^2 y + \sin^2 y = 1$ , and choose the sign BY THE RANGE: $\cos y \ge 0$ there.
5 Conclude with the domain: the formula holds for $-1 < x < 1$ , where the square root is positive.

Concluding sentence

“Let $y = \arcsin x$ , so $\sin y = x$ with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ . Differentiating, $\cos y \cdot y' = 1$ . On this interval $\cos y \ge 0$ , so $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ , and for $-1 < x < 1$ , $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ .”

The trap: Writing $\pm\sqrt{1 - x^2}$ and keeping the plus sign without a reason. For $\arctan$ there is no sign to discuss, since $\sec^2 y = 1 + \tan^2 y = 1 + x^2$ directly.

Marking: Typically 1 mark for the equation with its range, 1 for the implicit differentiation, 2 for the sign argument, 1 for the final formula with its domain.

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

Before differentiating, substitute the given point into the equation. A point that is not on the curve makes every slope that follows meaningless, and the check takes five seconds.

$(3, 1)$ on $2(x^2 + y^2)^2 = 25(x^2 - y^2)$ : $2 \cdot 100 = 200$ and $25 \cdot 8 = 200$ . A typo giving $(3, 2)$ would show $338 \ne 125$ at once.
The explicit branch agrees

When the curve can be solved for y near the point, differentiate that branch and compare. The two routes must give the same number.

$y = \sqrt{10 - x^2}$ has $y' = -\frac{x}{\sqrt{10 - x^2}}$ , which is $-\frac{1}{3}$ at $x = 1$ , as $-\frac{x}{y}$ gives at $(1, 3)$ .
Symmetry in y = x inverts slopes

If swapping x and y leaves the equation unchanged, the slopes at (a, b) and (b, a) must be reciprocal. The same test checks an inverse function.

On $x^3 + y^3 = 9xy$ : $\frac{4}{5}$ at $(2, 4)$ and $\frac{5}{4}$ at $(4, 2)$ ; for $f(x) = x^3 + x$ : $4 \cdot \frac{1}{4} = 1$ .
The sign of the inverse trigonometric derivative

arcsin and arctan increase, so their derivatives are positive; arccos decreases, so its derivative is negative. A sign that disagrees is a wrong sign argument.

$\frac{d}{dx}\arccos x$ at $x = \frac{1}{2}$ is $-\frac{2}{\sqrt 3} < 0$ , matching a graph that goes down.

The typical problem, taken apart

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

The curve $x^2 + 3xy + y^2 = 5$ passes through $P(1, 1)$ . Find the tangent and the normal lines at $P$ , and the value of $y''$ at $P$ .

No calculator. Every step must be justified as on a MATH 140 final.

The curve

x^2 + 3xy + y^2 = 5

has two branches, and

P(1, 1)

is on the upper one; the tangent and the normal are what the question asks for.

Step 1

Check: $1 + 3 + 1 = 5$ , so $P$ is on the curve. Differentiate both sides with respect to $x$ , product rule on $3xy$ and chain rule on $y^2$ : $2x + 3y + 3xy' + 2yy' = 0$ .

Why

The check is worth a mark and protects the rest. Naming the two rules shows the marker that the $y'$ factors did not appear by luck.

Step 2

Substitute $x = 1$ , $y = 1$ before isolating: $2 + 3 + 3y' + 2y' = 0$ , so $5y' = -5$ and $y' = -1$ .

Why

Only the slope at $P$ is asked, so the general formula $y' = -\frac{2x + 3y}{3x + 2y}$ is not needed: substituting first leaves a linear equation in one unknown.

Step 3

Tangent: $y - 1 = -1(x - 1)$ , that is $y = 2 - x$ . Normal, slope $1$ : $y - 1 = x - 1$ , that is $y = x$ .

Why

The normal is perpendicular to the tangent, so its slope is $-\frac{1}{y'}$ ; here the normal passes through the origin, a free check on the arithmetic.

Step 4

Differentiate the equation $2x + 3y + (3x + 2y)y' = 0$ once more, $y$ and $y'$ being functions of $x$ : $2 + 3y' + (3 + 2y')y' + (3x + 2y)y'' = 0$ .

Why

Differentiating the already differentiated equation avoids the quotient rule entirely. The product rule on $(3x + 2y)y'$ is where the $y''$ appears.

Step 5

At $P$ , with $y' = -1$ : $2 - 3 + (3 - 2)(-1) + 5y'' = 0$ , so $-2 + 5y'' = 0$ and $y'' = \frac{2}{5}$ .

Why

Numbers go in only AFTER the second differentiation: substituting $y' = -1$ first would freeze $y'$ as a constant and lose the term $(2y')y'$ .

Step 6

Verification with the explicit branch $y = \frac{-3x + \sqrt{5x^2 + 20}}{2}$ : at $x = 1$ , $y = \frac{-3 + 5}{2} = 1$ ; $y' = \frac{1}{2}\left(-3 + \frac{5x}{\sqrt{5x^2 + 20}}\right) = \frac{1}{2}(-3 + 1) = -1$ .

Why

The curve is quadratic in $y$ , so a branch can be written; comparing gives an independent check of the slope in two lines.

The conclusion, written out

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

The classic mistake on this problem: Replacing $y'$ by $-1$ in $2x + 3y + (3x + 2y)y' = 0$ and then differentiating: the equation becomes $2x + 3y - 3x - 2y = 0$ , a line, and $y''$ comes out as $0$ instead of $\frac{2}{5}$ .

Learn by heart

• Differentiate BOTH sides with respect to $x$ ; every $y$ -term leaves a factor $y'$ .
• $\frac{d}{dx}(xy) = y + xy'$ : a mixed term always needs the product rule.
• A slope belongs to a POINT $(x, y)$ of the curve, never to an $x$ alone.
• Horizontal: $N = 0$ , $D \ne 0$ ; vertical: $D = 0$ , $N \ne 0$ ; $\frac{0}{0}$ decides nothing.
• $y''$ : differentiate with $y$ still a function, then simplify with the equation of the curve.
• $(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}$ , $(\arccos x)' = -\frac{1}{\sqrt{1 - x^2}}$ , $(\arctan x)' = \frac{1}{1 + x^2}$ ; the RANGE fixes the sign.
• $(f^{-1})'(b) = \frac{1}{f'(a)}$ with $f(a) = b$ and $f'(a) \ne 0$ .

Frequently asked questions

How do you do implicit differentiation step by step?

Differentiate both sides of the equation with respect to x, treating y as a function of x, so that every term containing y gets a factor dy/dx from the chain rule and every mixed term like xy needs the product rule. Then move all the dy/dx terms to one side, factor dy/dx out, and divide. If only the slope at one point is asked, substitute the coordinates before isolating.

Why does my dy/dx answer still contain y?

Because an implicit curve can have several points above the same value of x, each with its own slope. The circle x squared plus y squared equals 10 has the points (1, 3) and (1, -3), with slopes -1/3 and 1/3. The y in the formula tells the points apart, so to get a number you substitute both coordinates of a point of the curve.

How do I find horizontal and vertical tangent lines of an implicit curve?

Write dy/dx as a fraction. For horizontal tangents, set the numerator to zero, solve that condition together with the equation of the curve, and keep the points where the denominator is not zero. For vertical tangents, do the same with the denominator, keeping the points where the numerator is not zero. If both vanish, the formula gives no conclusion.

How do you prove the derivative of arcsin x with implicit differentiation?

Write y = arcsin x as sin y = x with y between -pi/2 and pi/2. Differentiate: cos y times dy/dx = 1, so dy/dx = 1 / cos y. Since cos y is not negative on that interval, cos y equals the square root of 1 - x squared, and the derivative is 1 over the square root of 1 - x squared, for x strictly between -1 and 1.

How do I find the derivative of an inverse function at a point?

Find the number a such that f(a) = b, usually by inspection or from a table, then compute 1 divided by f prime of a. Do not evaluate f prime at b: b is an output of f, not an input. The formula needs f prime of a to be nonzero; where it is zero, the inverse has a vertical tangent.

Practise it

Corrected exercises: Implicit differentiation, MATH 140 at McGill

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 Logarithmic differentiation

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-implicit-differentiation. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: implicit differentiation and inverse trigonometric derivatives (MATH 140)

The essentials

y is a function you cannot see: every y-term leaves a y'

Inverse functions: the range fixes the sign, the right point fixes the value

The rules in table form

Differentiating each kind of term with respect to x

The mistakes that cost marks

1. Forgetting the factor y' on a y-term

2. Skipping the product rule on a mixed term

3. Giving a slope for an x instead of a point

4. Answering a horizontal-tangent question with a line

5. Reading a zero denominator as a vertical tangent

6. Differentiating y' as if y were a constant

7. Evaluating the inverse derivative at the wrong point

8. Leaving the sign of the square root to chance

9. Reading sin to the power minus 1 as a reciprocal

Which method to choose

Which move, by what the question asks

How the answer is expected to be written

Proving the derivative of an inverse trigonometric function

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The typical problem, taken apart

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

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Frequently asked questions

How do you do implicit differentiation step by step?

Why does my dy/dx answer still contain y?

How do I find horizontal and vertical tangent lines of an implicit curve?

How do you prove the derivative of arcsin x with implicit differentiation?

How do I find the derivative of an inverse function at a point?

Corrected exercises: Implicit differentiation, MATH 140 at McGill

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