Chapter 5 · Continuity and Differentiability

Exercise 5.3 · Q7: $\frac{dy}{dx}$ if $\sin^2y+\cos xy=\kappa$

Exercise 5.3Implicit differentiation · Section 5.3.2

The question

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

\sin^2 y + \cos xy = \kappa

Exam answer

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

\begin{aligned} &2\sin y\cos y\,\frac{dy}{dx} \\ &\quad - \sin xy\left(y + x\frac{dy}{dx}\right) = 0 \end{aligned}

Using $2\sin y\cos y = \sin 2y$ ⁠,

\begin{aligned} \Rightarrow\ &(\sin 2y - x\sin xy)\frac{dy}{dx} \\ &= y\sin xy \end{aligned}

$\therefore\ \dfrac{dy}{dx} = \dfrac{y\sin xy}{\sin 2y - x\sin xy}$ $(\sin 2y - x\sin xy \ne 0)$ ⁠.

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

The one idea this problem is built on

Here $\kappa$ is a constant, so the left side does not change along the curve and its derivative is $0$ ⁠. Both terms are compositions, so both need the chain rule: $\sin^2 y$ is "square of sine of $y$ ", and $\cos xy$ is "cosine of the product $xy$ ". Inside the second one, the product $xy$ needs the product rule. After that it is the usual collect, factor and divide.

Given, to find, and the plan

Given: $\sin^2 y + \cos xy = \kappa$ ⁠, where $\kappa$ is a constant and $y$ is a differentiable function of $x$ along the curve. ( $\cos xy$ means $\cos(xy)$ ⁠.)
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: Differentiate $\sin^2 y$ and $\cos(xy)$ by the chain rule; collect the $\frac{dy}{dx}$ terms; factor; divide where the bracket is not zero.

Basics you need

1Differentiating a relation

Both sides of a relation that holds along a piece of curve have equal derivatives there; a constant such as $\kappa$ has derivative $0$ ⁠. Lesson: implicit differentiation

2Chain rule

$\frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$ ⁠: derivative of the outer function at the inner one, times the derivative of the inner one (Theorem 4). Lesson: the chain rule

3Derivatives of sine and cosine; product rule

$(\sin t)' = \cos t$ ⁠, $(\cos t)' = -\sin t$ (Table 5.3); $(uv)' = u'v + uv'$ ⁠. Basics: derivative rules

4Double-angle formula

$2\sin y\cos y = \sin 2y$ (Class XI). Basics: trigonometric identities

Solution

Differentiate $\sin^2 y$

$\sin^2 y = (\sin y)^2$ ⁠. The outer function is $t^2$ ⁠, whose derivative is $2t$ ⁠; the inner function is $\sin y$ ⁠, whose derivative with respect to $x$ is $\cos y\,\frac{dy}{dx}$ (chain rule again, since $y$ depends on $x$ ⁠). So
$\begin{aligned} \frac{d}{dx}(\sin^2 y) &= 2\sin y\cdot\cos y\,\frac{dy}{dx} \\ &= \sin 2y\,\frac{dy}{dx}. \end{aligned}$
There are three layers: squaring, sine, and $y$ as a function of $x$ ⁠. Each layer contributes one factor. The double-angle formula only tidies the answer, and it matches the form used in the textbook answer.

Chain rule extended to three functions (Section 5.3.1); $2\sin y\cos y = \sin 2y$ ⁠. Basics: trigonometric identities
Differentiate $\cos(xy)$

The outer function is $\cos t$ ⁠, with derivative $-\sin t$ ⁠; the inner function is $t = xy$ ⁠, whose derivative is $1\cdot y + x\,\frac{dy}{dx}$ by the product rule. So
$\begin{aligned} \frac{d}{dx}\cos(xy) &= -\sin(xy)\left(y + x\frac{dy}{dx}\right) \\ &= -y\sin xy - x\sin xy\,\frac{dy}{dx}. \end{aligned}$
The derivative of the outer cosine is evaluated at the inner function $xy$ ⁠, not at $x$ or $y$ ⁠. The product rule is needed because both $x$ and $y$ change.

Chain rule, Theorem 4; product rule (2), Section 5.3; $(\cos t)' = -\sin t$ ⁠, Table 5.3. Basics: derivative rules
Put the pieces together

The derivative of the constant $\kappa$ is $0$ ⁠, so Steps 1 and 2 give
$\sin 2y\,\frac{dy}{dx} - y\sin xy - x\sin xy\,\frac{dy}{dx} = 0.$
Keep the $\frac{dy}{dx}$ terms on the left and move $-y\sin xy$ to the right:
$(\sin 2y - x\sin xy)\,\frac{dy}{dx} = y\sin xy.$
The constant $\kappa$ has dropped out: the slope formula is the same for every curve of the family, whatever the value of $\kappa$ ⁠.

Collect and factor. Basics: solving for dy/dx
Divide, where allowed

At points of the curve where $\sin 2y - x\sin xy \ne 0$ ⁠,
$\frac{dy}{dx} = \frac{y\sin xy}{\sin 2y - x\sin xy}.$
Where the bracket is zero we cannot divide. At such points Step 3 reads $0\cdot\frac{dy}{dx} = y\sin xy$ ⁠: if the right side is not zero no slope fits (a vertical tangent), and if it is zero the equation says nothing about the slope. So the formula is claimed only where the denominator is non-zero.

Result

\frac{dy}{dx} = \frac{y\sin xy}{\sin 2y - x\sin xy},

wherever $\sin 2y - x\sin xy \ne 0$ ⁠.

(Textbook answer: $\dfrac{y\sin xy}{\sin 2y - x\sin xy}$ ⁠, the same.)

Check it with numbers

Choose the curve through $(1, 0.5)$ ⁠: that needs $\kappa = \sin^2 0.5 + \cos 0.5 \approx 1.1074314$ ⁠. At that point:

numerator $y\sin xy = 0.5\sin 0.5 \approx 0.2397128$ ⁠;
denominator $\sin 2y - x\sin xy = \sin 1 - \sin 0.5$ ⁠, with $\sin 1 \approx 0.8414710$ and $\sin 0.5 \approx 0.4794255$ ⁠, so the denominator is about $0.3620454$ (to 7 places);
formula: $\frac{0.2397128}{0.3620454} \approx 0.6621068$ ⁠.

Directly: solving $\sin^2 y + \cos xy = 1.1074314\ldots$ for $y$ near $0.5$ at $x = 1 \pm 0.000001$ and taking the difference quotient gives $0.6621068$ ⁠. They agree.

Common mistakes

$\frac{d}{dx}(\sin^2 y) = 2\sin y\cos y$ ⁠. The last factor $\frac{dy}{dx}$ is missing; $y$ is a function of $x$ ⁠.
$\frac{d}{dx}\cos(xy) = -\sin(xy)$ ⁠. The derivative of the inner function $xy$ is missing.
$\frac{d}{dx}(xy) = x\frac{dy}{dx}$ or $= y$ ⁠. The product rule gives both terms: $y + x\frac{dy}{dx}$ ⁠.
Treating $\kappa$ as a variable. It is a constant, so its derivative is $0$ ⁠.

What to take away

Peel compositions layer by layer: each layer gives one factor in the chain rule.
A constant on one side of the relation disappears on differentiating, so the slope formula is shared by a whole family of curves.

Every fact used, and where it comes from

Fact	Used in	Source
Differentiating both sides of a relation; derivative of a constant is $0$	Step 3	Section 5.3.2, Examples 22 and 23 (Lesson: implicit differentiation)
Chain rule, also for three functions	Steps 1, 2	Theorem 4 and the extension after it, Section 5.3.1
$(\sin t)' = \cos t$ ⁠, $(\cos t)' = -\sin t$ ⁠, $(t^2)' = 2t$	Steps 1, 2	Table 5.3, Section 5.3
Product rule	Step 2	Rule (2), Section 5.3
$2\sin y\cos y = \sin 2y$	Step 1	Class XI, Trigonometric Functions
Numerical values	Check	Recomputed in the site's verification script

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Exercise 5.3 · Q7: dydx\frac{dy}{dx}dxdy​ if sin⁡2y+cos⁡xy=κ\sin^2y+\cos xy=\kappasin2y+cosxy=κ

Basics you need

Solution

Differentiate sin⁡2y\sin^2 ysin2y

Differentiate cos⁡(xy)\cos(xy)cos(xy)

Put the pieces together

Divide, where allowed

Every fact used, and where it comes from

Exercise 5.3 · Q7: $\frac{dy}{dx}$ if $\sin^2y+\cos xy=\kappa$

Differentiate $\sin^2 y$

Differentiate $\cos(xy)$