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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}

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Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

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