Chapter 5 · Continuity and Differentiability

Exercise 5.3 · Q4: $\frac{dy}{dx}$ if $xy+y^2=\tan x+y$

Exercise 5.3Implicit differentiation · Section 5.3.2

The question

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

xy + y^2 = \tan x + y

Exam answer

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

\begin{aligned} &y + x\frac{dy}{dx} + 2y\frac{dy}{dx} \\ &= \sec^2 x + \frac{dy}{dx} \end{aligned}

\begin{aligned} \Rightarrow\ &(x + 2y - 1)\frac{dy}{dx} \\ &= \sec^2 x - y \end{aligned}

$\therefore\ \dfrac{dy}{dx} = \dfrac{\sec^2 x - y}{x + 2y - 1}$ $(x + 2y - 1 \ne 0)$ ⁠.

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 $x$ ⁠. The term $xy$ is a product of two functions of $x$ (namely $x$ and $y$ ⁠), so it needs the product rule; $y^2$ needs the chain rule. Then $\frac{dy}{dx}$ appears in three places: collect, factor, divide.

Given, to find, and the plan

Given: $xy + y^2 = \tan x + y$ ⁠, with $y$ a function of $x$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: Differentiate term by term (product rule for $xy$ ⁠, chain rule for $y^2$ ⁠, Table 5.3 for $\tan x$ ⁠). Move all $\frac{dy}{dx}$ terms to the left, factor, divide, and state the conditions.

Basics you need

1Differentiating a relation

Both sides of a relation that holds on an interval have equal derivatives there (Section 5.3.2). Lesson: implicit differentiation

2Product rule

$(uv)' = u'v + uv'$ (rule (2), Section 5.3). With $u = x$ ⁠, $v = y$ ⁠: $\frac{d}{dx}(xy) = 1\cdot y + x\frac{dy}{dx}$ ⁠. Basics: derivative rules

3Chain rule on $y^2$

$\frac{d}{dx}(y^2) = 2y\,\frac{dy}{dx}$ ⁠. Lesson: the chain rule

4Derivative of $\tan x$

$(\tan x)' = \sec^2 x$ ⁠, valid where $\cos x \ne 0$ (Table 5.3). Basics: trigonometric identities

Solution

Differentiate each term with respect to $x$
$\begin{aligned} &\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) \\ &\quad = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y). \end{aligned}$
- $\frac{d}{dx}(xy) = y + x\,\frac{dy}{dx}$ (product rule).
- $\frac{d}{dx}(y^2) = 2y\,\frac{dy}{dx}$ (chain rule).
- $\frac{d}{dx}(\tan x) = \sec^2 x$ (Table 5.3).
- $\frac{d}{dx}(y) = \frac{dy}{dx}$ ⁠.
So
$\begin{aligned} &y + x\,\frac{dy}{dx} + 2y\,\frac{dy}{dx} \\ &\quad = \sec^2 x + \frac{dy}{dx}. \end{aligned}$
In $xy$ both factors change with $x$ ⁠, so its rate has two parts: $y$ times the rate of $x$ (which is $1$ ⁠) plus $x$ times the rate of $y$ ⁠. Writing just $x\frac{dy}{dx}$ or just $y$ would keep only one part.

Product rule (2) and Table 5.3, Section 5.3; chain rule, Theorem 4. Basics: derivative rules
Collect the $\frac{dy}{dx}$ terms

Subtract $\frac{dy}{dx}$ and $y$ from both sides:
$\begin{aligned} x\frac{dy}{dx} + 2y\frac{dy}{dx} - \frac{dy}{dx} &= \sec^2 x - y, \\ (x + 2y - 1)\,\frac{dy}{dx} &= \sec^2 x - y. \end{aligned}$
The $\frac{dy}{dx}$ on the right (coming from the term $y$ ⁠) must be brought over too; it becomes the $-1$ in the bracket. The term $y$ without a derivative moves to the right.

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

Wherever $x + 2y - 1 \ne 0$ (and $\cos x \ne 0$ ⁠, so that $\tan x$ is defined),
$\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}.$
The bracket is the divisor, so it must not be zero. The relation itself already requires $\cos x \ne 0$ ⁠, because it contains $\tan x$ ⁠.

Compare Example 23, where the book also states the condition that keeps the divisor non-zero. Lesson: implicit differentiation

Result

\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}, \quad x + 2y - 1 \ne 0.

(Textbook answer: $\dfrac{\sec^2 x - y}{x + 2y - 1}$ ⁠, the same.)

Check it with numbers

At $x = 0$ the relation reads $y^2 = y$ ⁠, so the curve passes through $(0, 0)$ and $(0, 1)$ ⁠. The formula gives

\begin{aligned} \text{at } (0, 0):\ \frac{dy}{dx} &= \frac{1 - 0}{0 + 0 - 1} = -1, \\ \text{at } (0, 1):\ \frac{dy}{dx} &= \frac{1 - 1}{0 + 2 - 1} = 0. \end{aligned}

For a direct check, solve the relation as a quadratic in $y$ ⁠: $y^2 + (x - 1)y - \tan x = 0$ ⁠, so

y = \frac{(1 - x) \mp \sqrt{(1 - x)^2 + 4\tan x}}{2}.

The branch with the minus sign passes through $(0, 0)$ ⁠; a difference quotient of it at $x = 0$ with $h = 0.0001$ gives $-1.0000000$ ⁠. The branch with the plus sign passes through $(0, 1)$ and gives $0.0000000$ ⁠. At $x = 0.5$ on the first branch, $y \approx -0.5302580$ ⁠; the formula gives $-1.1718588$ and the difference quotient gives $-1.1718588$ ⁠.

Common mistakes

$\frac{d}{dx}(xy) = x\frac{dy}{dx}$ or $= \frac{dy}{dx}$ ⁠. Both factors vary; the product rule gives $y + x\frac{dy}{dx}$ ⁠.
Forgetting the $\frac{dy}{dx}$ from the right-hand $y$ ⁠. It is what produces the $-1$ in the denominator.
Sign errors when moving terms. Collect carefully: the result is $(x + 2y - 1)y' = \sec^2 x - y$ ⁠.

What to take away

A product of $x$ and $y$ needs the product rule; a power of $y$ needs the chain rule; a lone $y$ gives $\frac{dy}{dx}$ ⁠.
Bring every $\frac{dy}{dx}$ term to one side, including those that started on the right.

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)
Product rule $(uv)' = u'v + uv'$	Step 1	Rule (2), Section 5.3
Chain rule	Step 1	Theorem 4, Section 5.3.1
$(\tan x)' = \sec^2 x$ ⁠, $(x^n)' = nx^{n-1}$	Step 1	Table 5.3, Section 5.3
Quadratic formula	Check	Class X, Quadratic Equations

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Exercise 5.3 · Q4: dydx\frac{dy}{dx}dxdy​ if xy+y2=tan⁡x+yxy+y^2=\tan x+yxy+y2=tanx+y

Basics you need

Solution

Differentiate each term with respect to xxx

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

Divide, where allowed

Every fact used, and where it comes from

Exercise 5.3 · Q4: $\frac{dy}{dx}$ if $xy+y^2=\tan x+y$

Differentiate each term with respect to $x$

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