# Calculus

## Implicit Differentiation

Sometimes functions cannot be defined explicitly in terms of one variable. For instance, the function y = f(x) = 2x3 + x + 1 is defined explicitly in terms of one variable, however, the function x3 + y3 = 9xy is not defined explicitly in terms of one variable because it is difficult to isolate for x or y. However, we can define the function implicitly in terms of a relation between x and y. In these cases, implicit differentiation is used to find $$\frac{dy}{dx}$$ or y’.

Implicit differentiation involves differentiating both sides of the equation with respect to one variable, (e.g., x) and treating the other variables (e.g., y) as a differentiable function with respect to x.

Let’s look at some examples to see what this means.

Example 1: Given x2 + y2 = 4, find $$\frac{dy}{dx}$$ .

Solution: Differentiate both sides of the equation with respect to x.

\begin{align} \frac{d}{dx}{x^2+y^2} &= \frac{d}{dx}4 \\ \frac{d}{dx}{x^2} + \frac{d}{dx}y^2 &=0 \\ 2x+2y \frac{dy}{dx} &=0 \\ 2y\frac{dy}{dx} &= -2x \\ \frac{dy}{dx} &= \frac{-2x}{2y} \\ \frac{dy}{dx} &= \frac{-x}{y} \end{align}

Note: Do not forget the $$\frac{dy}{dx}$$!

Notice when you differentiate x in respect to x, you get $$\frac{dx}{dx}$$. However, when you differentiate y in respect to x, you get $$\frac{dy}{dx}$$. This is what we call implicit differentiation and we mark it by keeping the $$\frac{dy}{dx}$$ every time a derivate of y in respect of x is performed.

Example 2: Given x3 + y3 = 9xy, find $$\frac{dy}{dx}$$ .

Solution: Differentiate both sides in respect to x

\begin{align} \frac{d}{dx} \left(x^3 + y^3\right) &= \frac{d}{dx} 9xy \\ \frac{d}{dx} x^3 + \frac{d}{dx} y^3 &= 9y + 9x \frac{dy}{dx} \\ 3x^2 + 3y^2\frac{dy}{dx} &= 9y + 9x\frac{dy}{dx} \end{align}

Divide both sides of equation by 3 and isolate for $$\frac{dy}{dx}$$,

\begin{align} x^2 + y^2 \frac{dy}{dx} &= 3y + 3x\frac{dy}{dx} \\ y^2 \frac{dy}{dx} - 3x \frac{dy}{dx} &= 3y -x^2 \\ \frac{dy}{dx} = \frac{3y-x^2}{y^2-3x} \end{align}

The variable you are differentiating in respect to determines the implicit differentiation. Sometimes you have multiple variables, but some are constant.

Example 3: The pressure $$p$$, volume $$V$$, and temperature $$T$$ of a gas are related by

$pV = n\left(RT+ap-bp/T\right)$

where $$a$$, $$b$$, $$n$$, and $$R$$, are constants. For constant $$V$$, find $$dp/dT$$.

Solution: Differentiate in respect to $$T$$. Then isolate for $$dp/dT$$.

\begin{align} \frac{d}{dT} pV &= \frac{d}{dT} \left(nRT+nap-\frac{nbp}{T}\right) \\ V \frac{dp}{dT} &= nR + na\frac{dp}{dT} - nb\frac{T-p\frac{dp}{dT}}{T^2} \\ V \frac{dp}{dT} - na\frac{dp}{dT} &= nR - \frac{nb}{T} + \frac{nbp}{T^2}\frac{dp}{dT} \\ V \frac{dp}{dT} - na\frac{dp}{dT} - \frac{nbp}{T^2}\frac{dp}{dT} &= nR - \frac{nb}{T} \\  \frac{dp}{dT}\left( V-na- \frac{nbp}{T^2}\right) &= nR - \frac{nb}{T} \\ \frac{dp}{dT}\left(\frac{VT^2-naT^2-nbp}{T^2}\right) &= \frac{nRT - nb}{T} \\ \frac{dp}{dT} &= \frac{nrT^2 - nbT}{VT^2-naT^2-nbp} \end{align}