Chapter 5 · Continuity and Differentiability

Exercise 5.3 · Q3: $\frac{dy}{dx}$ if $ax+by^2=\cos y$

Exercise 5.3Implicit differentiation · Section 5.3.2

The question

Find $\dfrac{dy}{dx}$ in the following:

ax + by^2 = \cos y

Exam answer

Differentiating both sides w.r.t. $x$ ⁠,

a + 2by\,\frac{dy}{dx} = -\sin y\,\frac{dy}{dx}

\Rightarrow\ (2by + \sin y)\frac{dy}{dx} = -a

$\therefore\ \dfrac{dy}{dx} = -\dfrac{a}{2by + \sin y}$ (where $2by + \sin y \ne 0$ ⁠).

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Differentiate both sides with respect to $x$ ⁠. Two terms are functions of $y$ ⁠, namely $y^2$ and $\cos y$ ⁠, and each one produces a factor $\frac{dy}{dx}$ by the chain rule. Collect those terms, factor out $\frac{dy}{dx}$ ⁠, and divide, noting where the divisor is zero.

Given, to find, and the plan

Given: $ax + by^2 = \cos y$ ⁠, where $a$ and $b$ are constants and $y$ is a function of $x$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: Differentiate each term; use the chain rule on $y^2$ and $\cos y$ ⁠; collect, factor, divide; state the condition for the division.

Basics you need

1Differentiating a relation

Both sides of a relation that holds for all $x$ in an interval have equal derivatives there (Section 5.3.2). Lesson: implicit differentiation

2Chain rule on functions of $y$

$\frac{d}{dx}(y^2) = 2y\,\frac{dy}{dx}$ and $\frac{d}{dx}(\cos y) = -\sin y\,\frac{dy}{dx}$ ⁠: the derivative with respect to $y$ ⁠, times $\frac{dy}{dx}$ ⁠. Lesson: the chain rule

3Constants

$a$ and $b$ are fixed numbers: $\frac{d}{dx}(ax) = a$ ⁠, and $b$ simply multiplies whatever $y^2$ gives. Basics: derivative rules

4Collecting and factoring

Gather the $\frac{dy}{dx}$ terms, factor, divide by the bracket where it is non-zero. Basics: solving for dy/dx

Solution

Differentiate each term with respect to $x$
$\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y).$
- $\frac{d}{dx}(ax) = a$ ⁠.
- $\frac{d}{dx}(by^2) = b\cdot 2y\,\frac{dy}{dx} = 2by\,\frac{dy}{dx}$ (outer function $t^2$ ⁠, inner function $y$ ⁠).
- $\frac{d}{dx}(\cos y) = -\sin y\,\frac{dy}{dx}$ (outer function $\cos t$ ⁠, inner function $y$ ⁠).
So
$a + 2by\,\frac{dy}{dx} = -\sin y\,\frac{dy}{dx}.$
$y^2$ and $\cos y$ are composites: something done to $y$ ⁠, which itself depends on $x$ ⁠. Each is differentiated with respect to $y$ (giving $2y$ and $-\sin y$ ⁠) and then multiplied by $\frac{dy}{dx}$ ⁠.

Chain rule, Theorem 4, Section 5.3.1; $(t^2)' = 2t$ ⁠, $(\cos t)' = -\sin t$ from Table 5.3. Lesson: the chain rule
Collect the $\frac{dy}{dx}$ terms

Add $\sin y\,\frac{dy}{dx}$ to both sides and subtract $a$ ⁠:
$\begin{aligned} 2by\,\frac{dy}{dx} + \sin y\,\frac{dy}{dx} &= -a, \\ (2by + \sin y)\,\frac{dy}{dx} &= -a. \end{aligned}$
All terms with $\frac{dy}{dx}$ go to one side and the rest to the other, so that $\frac{dy}{dx}$ can be factored out as a single unknown.

Factoring the common unknown. Basics: solving for dy/dx
Divide, where allowed

Wherever $2by + \sin y \ne 0$ ⁠,
$\frac{dy}{dx} = -\frac{a}{2by + \sin y}.$
Division by zero is not allowed, so the formula comes with the condition $2by + \sin y \ne 0$ ⁠, in the same way as Example 23 states a condition on $y$ ⁠.
What happens where $2by + \sin y = 0$ ⁠?

At such a point, Step 2 would read $0\cdot\frac{dy}{dx} = -a$ ⁠. If $a \ne 0$ this is impossible, so no finite $\frac{dy}{dx}$ exists there: the curve has a vertical tangent. For example, $y = 0$ gives $2b\cdot0 + \sin 0 = 0$ ⁠; the curve passes through $\left(\frac1a, 0\right)$ ⁠, and there $\frac{dx}{dy} = 0$ instead.

This is why the condition is not a technicality: it marks genuine points where $y$ does not change smoothly with $x$ ⁠. (The textbook answer, as usual, does not repeat the condition.)

Result

\frac{dy}{dx} = -\frac{a}{2by + \sin y}, \quad 2by + \sin y \ne 0.

(Textbook answer: $-\dfrac{a}{2by + \sin y}$ ⁠, the same.)

Check it with numbers

Take $a = 2$ ⁠, $b = 3$ ⁠, so the relation is $2x + 3y^2 = \cos y$ ⁠, and look at the point with $y = 1$ ⁠: $x = \frac{\cos 1 - 3}{2} \approx -1.2298488$ ⁠. The formula gives $-\frac{2}{6 + \sin 1} \approx -0.2923348$ ⁠. Moving along the curve from $y = 0.9999$ to $y = 1.0001$ (with $x = \frac{\cos y - 3y^2}{2}$ ⁠) and dividing the change in $y$ by the change in $x$ gives $-0.2923348$ ⁠. They agree.

Common mistakes

$\frac{d}{dx}(y^2) = 2y$ ⁠. The factor $\frac{dy}{dx}$ is missing.
$\frac{d}{dx}(\cos y) = \sin y\,\frac{dy}{dx}$ ⁠. The derivative of cosine is $-\sin$ ⁠; this minus sign is what puts $+\sin y$ next to $2by$ in the answer.
Treating $a$ or $b$ as variables. They are constants; $\frac{d}{dx}(ax) = a$ ⁠.

What to take away

Each term containing $y$ contributes (its $y$ -derivative) $\,\times\frac{dy}{dx}$ ⁠; terms in $x$ alone contribute ordinary derivatives.
The implicit derivative is valid where its denominator is non-zero; where it is zero the tangent is vertical (or the method gives no information).

Every fact used, and where it comes from

Fact	Used in	Source
Differentiating both sides of a relation	Step 1	Section 5.3.2, Examples 22 and 23 (Lesson: implicit differentiation)
Chain rule	Step 1	Theorem 4, Section 5.3.1
$(x^n)' = nx^{n-1}$ ⁠, $(\cos t)' = -\sin t$	Step 1	Table 5.3, Section 5.3
Division only by a non-zero quantity	Steps 3, 4	Algebra; compare the condition in Example 23

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Exercise 5.3 · Q3: dydx\frac{dy}{dx}dxdy​ if ax+by2=cos⁡yax+by^2=\cos yax+by2=cosy

Basics you need

Solution

Differentiate each term with respect to xxx

Collect the dydx\frac{dy}{dx}dxdy​ terms

Divide, where allowed

What happens where 2by+sin⁡y=02by + \sin y = 02by+siny=0⁠?

Every fact used, and where it comes from

Exercise 5.3 · Q3: $\frac{dy}{dx}$ if $ax+by^2=\cos y$

Differentiate each term with respect to $x$

Collect the $\frac{dy}{dx}$ terms

What happens where $2by + \sin y = 0$ ⁠?