Chapter 5 · Continuity and Differentiability

Exercise 5.3 · Q1: $\frac{dy}{dx}$ if $2x+3y=\sin x$

Exercise 5.3Implicit differentiation · Section 5.3.2

The question

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

2x + 3y = \sin x

Exam answer

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

2 + 3\frac{dy}{dx} = \cos x

\Rightarrow\ 3\frac{dy}{dx} = \cos x - 2

$\therefore\ \dfrac{dy}{dx} = \dfrac{\cos x - 2}{3}$ ⁠.

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

The one idea this problem is built on

If two expressions are equal for every $x$ ⁠, their derivatives with respect to $x$ are equal too. So we may differentiate both sides of the relation, term by term, remembering that $y$ is a function of $x$ whose derivative is the unknown $\frac{dy}{dx}$ ⁠. Then we solve for $\frac{dy}{dx}$ ⁠. Here $y$ can also be found explicitly, which gives an independent check.

Given, to find, and the plan

Given: $2x + 3y = \sin x$ ⁠, relating $y$ to $x$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: Differentiate both sides with respect to $x$ ⁠; solve the resulting equation for $\frac{dy}{dx}$ ⁠. Check by solving the relation for $y$ first.

Basics you need

1Differentiating a relation

If $L(x) = R(x)$ for all $x$ in an interval, then $\frac{d}{dx}L = \frac{d}{dx}R$ there. This is how Section 5.3.2 handles implicit functions (Examples 22 and 23). Lesson: implicit differentiation

2Sum and constant-multiple rules

$(u \pm v)' = u' \pm v'$ (rule (1), Section 5.3), and $(ky)' = k\,y'$ for a constant $k$ ⁠. Basics: derivative rules

3Standard derivatives

$(x)' = 1$ ⁠, $(\sin x)' = \cos x$ (Table 5.3). Basics: derivative rules

4Solving a linear equation for $\frac{dy}{dx}$

Treat $\frac{dy}{dx}$ as the unknown: move every other term to the other side, then divide by its coefficient. Basics: solving for dy/dx

Solution

Differentiate both sides with respect to $x$
$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x).$
Term by term: $\frac{d}{dx}(2x) = 2$ ⁠, $\frac{d}{dx}(3y) = 3\frac{dy}{dx}$ ⁠, $\frac{d}{dx}(\sin x) = \cos x$ ⁠. So
$2 + 3\frac{dy}{dx} = \cos x.$
The relation holds for every $x$ ⁠, so both sides are the same function of $x$ and have the same derivative. The term $3y$ is $3$ times an unknown function of $x$ ⁠; its derivative is $3$ times the unknown derivative $\frac{dy}{dx}$ ⁠. We do not need to know $y$ to write this.

Sum rule (1), Section 5.3; constant-multiple rule; Table 5.3. Lesson: implicit differentiation
Solve for $\frac{dy}{dx}$
$\begin{aligned} 3\frac{dy}{dx} &= \cos x - 2, \\ \frac{dy}{dx} &= \frac{\cos x - 2}{3}. \end{aligned}$
Subtract $2$ from both sides, then divide by $3$ ⁠, the coefficient of $\frac{dy}{dx}$ ⁠. Dividing by $3$ is always allowed, so the formula holds for every real $x$ ⁠.

Linear equation in the unknown $\frac{dy}{dx}$ ⁠. Basics: solving for dy/dx
Check by solving for $y$ first

Here $y$ can be isolated: $3y = \sin x - 2x$ ⁠, so $y = \dfrac{\sin x - 2x}{3}$ ⁠. Differentiating directly,
$\frac{dy}{dx} = \frac{\cos x - 2}{3},$
the same answer.

This is the comparison the book makes in Example 22: when $y$ can be written explicitly, both routes must agree. The implicit route is the one that still works when $y$ cannot be isolated (as in the next question).

Explicit and implicit functions, Section 5.3.2. Lesson: implicit differentiation
A remark on the sign

Since $\cos x \le 1$ ⁠, the numerator $\cos x - 2 \le -1$ ⁠, so $\frac{dy}{dx} \le -\frac13$ for every $x$ ⁠: $y$ is a strictly decreasing function of $x$ ⁠.

Reading a derivative this way is a useful sanity check: $y = \frac{\sin x}{3} - \frac{2x}{3}$ is a steady downward line $-\frac{2x}{3}$ with a small wobble $\frac{\sin x}{3}$ that is never steep enough to turn it upward.

A function whose derivative is negative throughout an interval is decreasing there: this is Theorem 1, Section 6.3 of Chapter 6, used here ahead of time and not needed for the answer. Basics: slope at a point

Result

\frac{dy}{dx} = \frac{\cos x - 2}{3}\quad\text{for every real } x.

(Textbook answer: $\dfrac{\cos x - 2}{3}$ ⁠, the same.)

Check it with numbers

At $x = 0$ ⁠: the formula gives $\frac{1 - 2}{3} = -\frac13 \approx -0.3333333$ ⁠. A difference quotient of $y = \frac{\sin x - 2x}{3}$ with $h = 0.0001$ on each side of $0$ gives $-0.3333333$ ⁠. At $x = \frac{\pi}{3}$ ⁠: formula $\frac{0.5 - 2}{3} = -0.5$ ⁠; difference quotient $-0.5000000$ ⁠.

Common mistakes

Differentiating $3y$ as $3$ ⁠. $y$ is not $x$ ⁠; its derivative is the unknown $\frac{dy}{dx}$ ⁠, so $\frac{d}{dx}(3y) = 3\frac{dy}{dx}$ ⁠.
Differentiating $2x$ as $2x$ or $0$ ⁠. $(2x)' = 2$ ⁠.
Sign slip when moving the $2$ ⁠. $2 + 3y' = \cos x$ gives $3y' = \cos x - 2$ ⁠, not $2 - \cos x$ ⁠.

What to take away

Implicit differentiation: differentiate every term of the relation with respect to $x$ ⁠, writing $\frac{dy}{dx}$ wherever $y$ is differentiated, then solve for $\frac{dy}{dx}$ ⁠.
When $y$ can be isolated, doing so gives a free check.

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)
Sum rule; constant-multiple rule	Steps 1, 3	Rule (1), Section 5.3
$(x)' = 1$ ⁠, $(\sin x)' = \cos x$	Steps 1, 3	Table 5.3, Section 5.3
Explicit versus implicit functions	Step 3	Section 5.3.2
$-1 \le \cos x \le 1$	Step 4	Class XI, Trigonometric Functions
Negative derivative on an interval means decreasing	Step 4	Theorem 1, Section 6.3 (Chapter 6)

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Exercise 5.3 · Q1: dydx\frac{dy}{dx}dxdy​ if 2x+3y=sin⁡x2x+3y=\sin x2x+3y=sinx

Basics you need

Solution

Differentiate both sides with respect to xxx

Solve for dydx\frac{dy}{dx}dxdy​

Check by solving for yyy first

A remark on the sign

Every fact used, and where it comes from

Exercise 5.3 · Q1: $\frac{dy}{dx}$ if $2x+3y=\sin x$

Differentiate both sides with respect to $x$

Solve for $\frac{dy}{dx}$

Check by solving for $y$ first