Chapter 5 · Continuity and Differentiability

Example 22: $\frac{dy}{dx}$ if $x-y=\pi$ ⁠, two ways

ExamplesImplicit differentiation · Section 5.3.2

The question

Find $\dfrac{dy}{dx}$ if $x - y = \pi$ ⁠.

Exam answer

Differentiating both sides of $x - y = \pi$ w.r.t. $x$ ⁠,

\frac{d}{dx}(x) - \frac{dy}{dx} = \frac{d}{dx}(\pi)

\Rightarrow\ 1 - \frac{dy}{dx} = 0

$\therefore\ \dfrac{dy}{dx} = 1$ ⁠.

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

The one idea this problem is built on

When an equation ties $y$ to $x$ ⁠, we can find $\frac{dy}{dx}$ without first solving for $y$ ⁠: differentiate both sides of the equation with respect to $x$ , treating $y$ as an unknown function of $x$ ⁠. This is implicit differentiation. Here we can also solve for $y$ ⁠, so we do it both ways and see that they agree. That agreement is what makes the direct method trustworthy for harder relations such as Example 23.

Given, to find, and the plan

Given: The relation $x - y = \pi$ between $x$ and $y$ ⁠, for all real $x$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: Method 1: solve for $y$ (explicit form) and differentiate. Method 2: differentiate the relation as it stands (implicit form). Compare.

Basics you need

1Explicit and implicit

$y = x - \pi$ gives $y$ explicitly: a formula in $x$ alone. $x - y = \pi$ gives the same $y$ implicitly: $y$ is fixed by $x$ ⁠, but you have to unpick the equation to see the formula. Basics: solving for dy/dx

2Derivative of a constant and of $x$

A constant function (such as the one with value $\pi$ everywhere) has derivative $0$ ⁠: its graph is flat. And $\frac{d}{dx}(x) = 1$ (power rule with $n = 1$ ⁠). Basics: derivative rules

3Difference rule

$(u - v)' = u' - v'$ ⁠: the derivative of a difference is the difference of the derivatives (algebra of derivatives, Section 5.3). Basics: slope at a point

Solution

Method 1: solve for $y$ ⁠, then differentiate

Add $y$ to both sides and subtract $\pi$ ⁠:
$x - y = \pi \iff y = x - \pi.$
Now differentiate term by term:
$\frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(\pi) = 1 - 0 = 1.$
The two equations $x - y = \pi$ and $y = x - \pi$ are equivalent (each step can be reversed), so they describe exactly the same $y$ ⁠. Once $y$ is an explicit formula in $x$ ⁠, ordinary rules apply. $\pi \approx 3.14159$ is just a number, so it differentiates to $0$ ⁠.

Difference rule (algebra of derivatives, Section 5.3), $\frac{d}{dx}(x) = 1$ (Table 5.3 with $n = 1$ ⁠) and derivative of a constant $= 0$ (Class XI, Limits and Derivatives). Basics: derivative rules
Method 2: differentiate the relation directly

Think of $y$ as some function of $x$ ⁠, $y = y(x)$ ⁠. The relation says the function $x - y(x)$ equals $\pi$ at every $x$ ⁠. Two functions that are equal at every $x$ have equal derivatives, so
$\frac{d}{dx}(x - y) = \frac{d}{dx}(\pi).$
The left side, by the difference rule, is $\frac{d}{dx}(x) - \frac{dy}{dx} = 1 - \frac{dy}{dx}$ ⁠. The right side is the derivative of the constant function with value $\pi$ ⁠, which is $0$ ⁠. So
$1 - \frac{dy}{dx} = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = 1.$
This is the key move of implicit differentiation, and it needs the relation to hold for all $x$ in an interval, not just at one point. Two functions equal everywhere on an interval have the same graph there, hence the same slope everywhere. We never needed the formula for $y$ ⁠: $\frac{dy}{dx}$ appeared as an unknown and we solved for it.

The book stresses that $\frac{d\pi}{dx}$ means differentiating "the constant function taking value $\pi$ everywhere", which gives $0$ ⁠. Lesson: implicit differentiation
Compare, and see what the answer means

Both methods give $\frac{dy}{dx} = 1$ ⁠. The graph of $x - y = \pi$ is the straight line $y = x - \pi$ ⁠, which rises 1 unit for each 1 unit to the right:

The line $x - y = \pi$ (equal scales on both axes). It crosses the axes at $(\pi, 0)$ and $(0, -\pi)$ ⁠. Its slope is $\frac{\text{rise}}{\text{run}} = \frac{1}{1} = 1$ everywhere, matching $\frac{dy}{dx} = 1$ ⁠.

A straight line has the same slope at every point, so $\frac{dy}{dx}$ is a constant. The value of the constant $\pi$ only shifts the line; it does not tilt it. Any relation $x - y = c$ gives $\frac{dy}{dx} = 1$ ⁠.

The derivative at a point is the slope of the tangent there; for a straight line the tangent is the line itself. Basics: slope at a point

Result

$\dfrac{dy}{dx} = 1$ for every real $x$ ⁠. (The book gets $\frac{dy}{dx} = \frac{dx}{dx} = 1$ by both methods.)

Check it with numbers

At $x = 4$ ⁠: $y = 4 - \pi = 0.8584073\ldots$ ⁠. At $x = 4.001$ ⁠: $y = 0.8594073\ldots$ ⁠. The change in $y$ is $0.001$ ⁠, equal to the change in $x$ ⁠, so $\frac{\Delta y}{\Delta x} = 1$ exactly (no rounding error, because the graph is a straight line).

Common mistakes

Treating $\frac{d\pi}{dx}$ as something other than 0. $\pi$ is a fixed number, not a variable.
Writing $\frac{d}{dx}(y) = 1$ ⁠. The derivative of $y$ with respect to $x$ is the unknown $\frac{dy}{dx}$ ⁠; only $\frac{d}{dx}(x) = 1$ ⁠.
Sign slip. $\frac{d}{dx}(x - y) = 1 - \frac{dy}{dx}$ ⁠, with a minus sign in front of $\frac{dy}{dx}$ ⁠.

What to take away

Implicit differentiation: differentiate every term of the relation with respect to $x$ ⁠, write $\frac{dy}{dx}$ wherever $y$ is differentiated, then solve for $\frac{dy}{dx}$ ⁠.
It is valid because the relation holds for all $x$ in an interval, so both sides are the same function.
When you can solve for $y$ ⁠, the explicit method is a good check on the implicit one.

Every fact used, and where it comes from

Fact	Used in	Source
$(u-v)' = u' - v'$ ⁠; $\frac{d}{dx}x = 1$ ⁠; constant $\to 0$	Steps 1, 2	Algebra of derivatives and Table 5.3, Section 5.3; derivative of a constant, Class XI (Limits and Derivatives)
Explicit and implicit functions	Steps 1, 2	Section 5.3.2 (Lesson: implicit differentiation)
Equal functions on an interval have equal derivatives	Step 2	Definition of the derivative as a limit, Section 5.3
Slope of a straight line	Step 3	Coordinate geometry (Class XI)

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Example 22: dydx\frac{dy}{dx}dxdy​ if x−y=πx-y=\pix−y=π⁠, two ways

Basics you need

Solution

Method 1: solve for yyy⁠, then differentiate

Method 2: differentiate the relation directly

Compare, and see what the answer means

Every fact used, and where it comes from

Example 22: $\frac{dy}{dx}$ if $x-y=\pi$ ⁠, two ways

Method 1: solve for $y$ ⁠, then differentiate