Implicit differentiation and inverse trigonometric functions

When a curve is described by an equation you cannot solve for $y$ , you differentiate both sides as they stand and treat $y$ as a function of $x$ . The chain rule does the work, and the result is a formula for the slope in terms of both coordinates. The same method gives the derivatives of the inverse trigonometric functions, and of any inverse function at a point.

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In this topic

Taught: Before the midterm
Questions: 30
Total marks: 85
Suggested time: 128 min
Difficulty: Introductory to Capstone
Show the 9 hardest

Key results

Implicit differentiation: $\frac{d}{dx}\,g(y) = g'(y)\,\frac{dy}{dx}$
When $y$ is a differentiable function of $x$ and $g$ is differentiable at $y$ .
The derivative of an inverse: $\bigl(f^{-1}\bigr)'(b) = \frac{1}{f'\bigl(f^{-1}(b)\bigr)}$
When $f$ is continuous and one-to-one on an open interval containing $f^{-1}(b)$ , differentiable at $f^{-1}(b)$ , and $f'\bigl(f^{-1}(b)\bigr) \neq 0$ .
Inverse trigonometric ranges: $-\tfrac{\pi}{2} \le \arcsin x \le \tfrac{\pi}{2}, \qquad 0 \le \arccos x \le \pi, \qquad -\tfrac{\pi}{2} < \arctan x < \tfrac{\pi}{2}$
For $-1 \le x \le 1$ in the first two and every $x$ in the third, so $\arcsin(\sin\theta) = \theta$ only when $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$ .
Arcsine and arccosine: $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^{2}}}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^{2}}}$
For $-1 < x < 1$ .
Arctangent: $\frac{d}{dx}\arctan x = \frac{1}{1 + x^{2}}$

Read the theoryUBC Math 100 §4.5 Implicit differentiationUBC Math 100 §4.7 Inverse trig functions and their derivativesCLP-1 §2.11 Implicit differentiationCLP-1 §2.12 Inverse trigonometric functions

Worked example

Standard3 marks · 4 min

The point $(1, 2)$ lies on the curve below. Find $\dfrac{dy}{dx}$ at that point.

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

Strategy

Differentiate the equation term by term with respect to x, remembering that y depends on x. Then gather the dy/dx terms on one side and solve, substituting the point only at the end.

Working

$\frac{d}{dx}\bigl(x^{2}\bigr) + \frac{d}{dx}(xy) + \frac{d}{dx}\bigl(y^{2}\bigr) = \frac{d}{dx}(7)$
Differentiate both sides. The right-hand side is a constant, so it becomes 0.
$2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0$
The product rule handles xy; the chain rule turns y² into 2y·dy/dx.
$\frac{dy}{dx}\bigl(x + 2y\bigr) = -(2x + y)$
Collect every dy/dx term on one side and factor it out.
$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$
Divide. This is the slope at any point on the curve.
$\left.\frac{dy}{dx}\right|_{(1,\,2)} = -\frac{2(1) + 2}{1 + 2(2)} = -\frac{4}{5}$
Substitute both coordinates. Implicit derivatives normally need x and y.

Answer

\left.\frac{dy}{dx}\right|_{(1,\,2)} = -\frac{4}{5}

Check

At x = 1.01 the curve’s y is about 1.9920, a drop of about 0.008 from 2 over a step of 0.01, a slope of −0.8.

Where this goes wrong

Differentiating xy as just y, or as just x·dy/dx. It is a product of two functions of x, so both terms appear.

On the paper

Differentiate both sides, attach $\frac{dy}{dx}$ to every term in $y$ as the chain rule requires, and solve for $\frac{dy}{dx}$ before substituting the point.

Leans on

The product rule, from Differentiation rules

Your turn

Standard3 marks · 4 min

The point $(2, 1)$ lies on the curve below. Find $\dfrac{dy}{dx}$ at that point.

x^{3} + \sin(\pi y) + y = 9

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Practice questions

5 of the 30, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory2 marks · 3 min
The point $(3, 4)$ lies on the circle $x^{2} + y^{2} = 25$ . Use implicit differentiation to find the slope of the tangent line there.
Strategy
Differentiate the relation as it stands, solve for y′, and substitute the coordinates of the point. The result agrees with geometry: the tangent to a circle is perpendicular to the radius.
Working
$2x + 2y\,y' = 0 \quad\Longrightarrow\quad y' = -\frac{x}{y}$
Implicit differentiation; the constant 25 differentiates to 0.
$y'\big|_{(3, 4)} = -\frac{3}{4}$
The radius to (3, 4) has slope 4/3; the tangent is perpendicular, slope −3/4.
Answer
$-\frac{3}{4}$
The answer is negative three quarters.
Check
The circle passes through (3.01, 3.9925), where y has dropped by about 0.0075 for a step of 0.01 in x, a slope of about −0.75.
Where this goes wrong
Differentiating y² as 2y, leaving out y′. Then 2x + 2y = 0 has no derivative in it to solve for, and it is false at (3, 4). The chain-rule factor y′ is what makes the differentiated equation about slopes.
Leans on
The chain rule, from The chain rule and logarithms
Try a similar question
The point $(1, 2)$ lies on the curve below. Find $\dfrac{dy}{dx}$ at that point. $x^{2} + xy + y^{2} = 7$
Answer this question
Introductory2 marks · 3 min
Let $f(x) = \arcsin\!\left(\dfrac{x}{3}\right)$ . Find $f'\!\left(\tfrac{3}{2}\right)$ .
Strategy
The derivative of arcsin is a standard result; the argument x/3 brings in a chain-rule factor of 1/3.
Working
$f'(x) = \frac{1}{\sqrt{1 - (x/3)^{2}}}\cdot\frac{1}{3}$
Derivative of arcsin at the inside, times the inner derivative.
$f'\!\left(\tfrac{3}{2}\right) = \frac{1}{\sqrt{1 - \frac{1}{4}}}\cdot\frac{1}{3} = \frac{2}{\sqrt{3}}\cdot\frac{1}{3} = \frac{2}{3\sqrt{3}} \approx 0.385$
At x = 3/2 the inside is 1/2, and √(3/4) = √3/2.
Answer
$f'\!\left(\frac{3}{2}\right) = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$
The answer is 2 over 3 root 3, about 0.385.
Check
The inverse, x = 3 sin y, has slope 3 cos(π/6) = 3√3/2 ≈ 2.598 at y = π/6, and 1/2.598 ≈ 0.385.
Where this goes wrong
Forgetting the inner derivative and answering 2/√3. The inside is x/3, so the chain rule multiplies by 1/3.
Leans on
The chain rule, from The chain rule and logarithms
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Let $g(x) = \bigl(1 + x^{2}\bigr)\arctan x$ . Find $g'(1)$ .
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Standard2 marks · 3 min
Given $\sin(y) + x^{2} = y$ , which expression is $\dfrac{dy}{dx}$ ?
Strategy
Differentiate each side with respect to x, letting the chain rule attach dy/dx to every term built from y, then isolate dy/dx.
Working
$\cos(y)\frac{dy}{dx} + 2x = \frac{dy}{dx}$
Both sin(y) and the bare y on the right produce a dy/dx factor.
$2x = \frac{dy}{dx} - \cos(y)\frac{dy}{dx}$
Move the dy/dx terms to the same side.
$2x = \frac{dy}{dx}\bigl(1 - \cos y\bigr)$
Factor dy/dx out of both terms.
$\frac{dy}{dx} = \frac{2x}{1 - \cos y}$
Divide by the bracket.
Answer
$\frac{dy}{dx} = \frac{2x}{1 - \cos y}$
The answer is 2x/(1 − cos y).
Check
With the curve written as x² = y − sin y, differentiating in y gives 2x · dx/dy = 1 − cos y, and flipping dx/dy gives 2x/(1 − cos y).
Where this goes wrong
Differentiating sin(y) as cos(y) and forgetting dy/dx, which leaves nothing to solve for. Every function of y contributes a dy/dx factor when you differentiate with respect to x.
Leans on
Chain rule with an unknown inner function, from The chain rule and logarithms
Try a similar question
The point $(1, 2)$ lies on the curve below. Find $\dfrac{dy}{dx}$ at that point. $x^{2} + xy + y^{2} = 7$
Answer this question
Challenging4 marks · 6 min
Verify that the point $(2, 4)$ lies on the curve below, then find the equation of the tangent line there.
$x^{3} + y^{3} = 9xy$
Give the tangent line in any correct form, then compare against the model solution.
Strategy
Confirm the point is on the curve, differentiate implicitly to get a slope formula in x and y, evaluate it at the point, and assemble the point-slope form.
Working
$2^{3} + 4^{3} = 8 + 64 = 72, \qquad 9(2)(4) = 72$
Both sides agree, so the point does lie on the curve.
$3x^{2} + 3y^{2}\frac{dy}{dx} = 9y + 9x\frac{dy}{dx}$
Differentiate both sides. The right-hand side is a product, so it splits into two terms.
$\frac{dy}{dx}\bigl(3y^{2} - 9x\bigr) = 9y - 3x^{2}$
Move every dy/dx term to the left and everything else to the right.
$\frac{dy}{dx} = \frac{9y - 3x^{2}}{3y^{2} - 9x} = \frac{3y - x^{2}}{y^{2} - 3x}$
Divide, then cancel the common factor of 3.
$\left.\frac{dy}{dx}\right|_{(2,\,4)} = \frac{3(4) - 4}{16 - 6} = \frac{8}{10} = \frac{4}{5}$
Substitute the point to get the numerical slope.
$y - 4 = \frac{4}{5}(x - 2)$
Point-slope form. Equivalently y = (4/5)x + 12/5.
Answer
$y - 4 = \frac{4}{5}(x - 2)$
The answer is y − 4 = (4/5)(x − 2), equivalently y = 0.8x + 2.4.
What earns the marks
Checks the point satisfies the equation: 8 + 64 = 72 and 9(2)(4) = 72.
Differentiates implicitly, applying the product rule to the right-hand side.
Solves for dy/dx and substitutes both coordinates to get a slope of 4/5.
States a correct equation for the tangent line, such as y − 4 = (4/5)(x − 2).
Check
The curve also passes through about (2.01, 4.0080), a rise of about 0.008 over a step of 0.01, a slope of about 0.8 = 4/5.
Where this goes wrong
Differentiating 9xy as 9·dy/dx, treating x as a constant. It is a genuine product, so it differentiates to 9y + 9x·dy/dx.
Leans on
Equation of a tangent line, from The definition of the derivative
Try a similar question
The point $(1, 2)$ lies on the curve below. Find $\dfrac{dy}{dx}$ at that point. $x^{2} + xy + y^{2} = 7$
Answer this question
Capstone5 marks · 9 min
The equation below defines $y$ as a differentiable function of $x$ for every real $x$ . Find $\dfrac{d^{2}y}{dx^{2}}$ at $x = 1 + \dfrac{\pi}{4}$ .
$y + \arctan y = x$
Strategy
First find the point: arctan 1 = π/4 gives y = 1, and it is the only y because y + arctan y is increasing. Implicit differentiation then gives y′ as a formula in y, and differentiating that formula again needs the chain rule once more, because y still depends on x; the numbers go in only at the end.
Working
$x = 1 + \tfrac{\pi}{4}:\qquad y + \arctan y = 1 + \tfrac{\pi}{4} \quad\Longrightarrow\quad y = 1$
arctan 1 = π/4, so y = 1 works; it is the only solution, because y + arctan y is increasing.
$y' + \frac{y'}{1 + y^{2}} = 1$
Differentiate both sides; the chain rule gives (arctan y)′ = y′/(1 + y²).
$y' = \frac{1}{1 + \frac{1}{1 + y^{2}}} = \frac{1 + y^{2}}{2 + y^{2}}$
Factor out y′. The bracket 1 + 1/(1 + y²) is never 0, so y′ exists at every point.
$y = 1:\qquad y' = \frac{1 + 1}{2 + 1} = \frac{2}{3}$
The slope at the point.
$y'' = \frac{2y\,y'\,(2 + y^{2}) - (1 + y^{2})\cdot 2y\,y'}{(2 + y^{2})^{2}} = \frac{2y\,y'}{(2 + y^{2})^{2}}$
The quotient rule, with (y²)′ = 2y y′; the y³ terms cancel.
$y'' = \frac{2\cdot 1\cdot \frac{2}{3}}{3^{2}} = \frac{4}{27}$
Now substitute y = 1 and y′ = 2/3.
Answer
$\left.\frac{d^{2}y}{dx^{2}}\right|_{x = 1 + \pi/4} = \frac{4}{27}$
The answer is four twenty-sevenths, about 0.148.
Check
Solving y + arctan y = x numerically, y′ = (1 + y²)/(2 + y²) rises from about 0.6519 at x = 1.6854 to 0.6815 at x = 1.8854, a rate of about 0.148.
Where this goes wrong
Treating y as a constant when differentiating y′ = (1 + y²)/(2 + y²) a second time, which gives 0. y changes with x, so the derivative of y² is 2y y′; substitute y = 1 and y′ = 2/3 only after differentiating, and remember that 1 + π/4 is the value of x, not of y.
Leans on
The chain rule, from The chain rule and logarithms
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The point $(-1, 2)$ lies on the curve below. Find $\dfrac{d^{2}y}{dx^{2}}$ at that point. $x^{3} + y^{3} = 7$
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Needs
The chain rule and logarithms
Compositions, nested rules, the natural logarithm and logarithmic differentiation.
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Translating a moving situation into an equation, then differentiating.
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Key results

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Working

Answer

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Practice questions

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Check

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The chain rule and logarithms

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