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Implicit Differentiation

In this section we learn how to perform implicit differentiation.
This technique will allow us to differentiate expressions in which \(y\) is not explicitly defined as a function of \(x\).
For instance, expressions like: \[3 y^2x - sin(y)+4x^2=3\] The method, or technique for differentiating expressions in which \(y\) is implicitly defined is shown here.

Method - Implicit Differentiation

To differentiate functions in which \(y\) is implicitly defined, two things should be kept in mind:

  • \(y\) is a function of \(x\).
    In fact when we start learning implicit differentiation it's often useful to replace every \(y\) we see by \(y(x)\).

  • The chain rule: \(\begin{pmatrix} g\begin{bmatrix} h(x)\end{bmatrix}\end{pmatrix}' = h'(x)\times g'\begin{pmatrix}h(x) \end{pmatrix}\).
Consequently, to differentiate composite functions involing \(y\), we use the chain rule: \[\begin{pmatrix}g\begin{bmatrix}y(x)\end{bmatrix}\end{pmatrix}' = y'(x)\times g'\begin{bmatrix}y(x)\end{bmatrix}\] We'll usually write this as: \[\begin{pmatrix}g\begin{bmatrix}y\end{bmatrix}\end{pmatrix}' = \frac{dy}{dx}\times g'\begin{bmatrix}y\end{bmatrix}\]


Differentiate the expression that follows with respect to \(x\): \[3y^2-x^2=8\]


We can re-write \(3y^2-x^2=8\) highlighting the fact that \(y\) is a function of \(x\), that's: \[3y(x)^2-x^2=8\] The key when differentiating this is to treat the expression \(y(x)^2\) in exactly the same way that we would treat composite expressions like \(\begin{pmatrix}sin(x)\end{pmatrix}^2\), or \(\begin{pmatrix}x^2-1\end{pmatrix}^2\), ... .

Using the chain rule, we find that the derivative of \(y(x)^2\) is \(2y'(x).y(x)\), that's: \[\begin{pmatrix}y(x)^2\end{pmatrix}' = 2y'(x)y(x)\]
We tend to write this in the form: \[\begin{pmatrix}y^2\end{pmatrix}' = 2\frac{dy}{dx}y\]

We differentiate the remaining terms of the expression the usual way.
So differentiating both sides of \(3y^2-x^2=8\) with respect to \(x\): \[3\times 2\frac{dy}{dx}y - 2x = 0\] That's: \[6\frac{dy}{dx}y - 2x = 0\] And that would be the final answer.

We could also go further and find an expression for \(\frac{dy}{dx}\). To do so we rearrange the result we just found, to make \(\frac{dy}{dx}\) the subject: \[6\frac{dy}{dx}y = 2x\] That's: \[\frac{dy}{dx} = \frac{2x}{6y}\] Finally: \[\frac{dy}{dx} = \frac{x}{3y}\]


Exercise 1

  1. Given \(y^2-x=4\), find an expression for \(\frac{dy}{dx}\).
  2. Given \(x^3+y^3 = 5\), find an expression for \(\frac{dy}{dx}\).
  3. Given \(y=2x.sin(y)\), find an expression for \(\frac{dy}{dx}\).
  4. Given \(4x.\sqrt{sin(y)}-5y^2=1\), find an expression for \(\frac{dy}{dx}\).

Answers Without Working

  1. Given \(y^2-x=4\), we find: \[\frac{dy}{dx} = \frac{1}{2y}\]
  2. Given \(x^3+y^3 = 5\), we find: \[\frac{dy}{dx} = -\frac{x^2}{y^2}\]
  3. Given \(y=2x.sin(y)\), we find: \[\frac{dy}{dx} = \frac{2.sin(y)}{1-2x.cos(y)}\]
  4. Given \(4x.\sqrt{sin(y)}-5y^2=1\), we find: \[\frac{dy}{dx} = \frac{2.sin(y)}{5y.\sqrt{sin(y)}-x.cos(y)}\]

Second and Higher Derivatives

At times we'll be required to find an expression for the second derivative \(\frac{d^2y}{dx^2}\) (or higher).

To do this we "simply" differentiate the expression, with respect to \(x\), twice and rearrange to make \(\frac{d^2y}{dx^2}\) the subject.


Given the expression: \[3x^2+y^2 = 7\]

  1. \(\frac{dy}{dx}\)
  2. \(\frac{d^2y}{dx^2}\)


We start by differentiating with respect to \(x\): \[3x^2+y^2 = 7\] becomes: \[6x+2\frac{dy}{dx}y = 0\]

  1. To find \(\frac{dy}{dx}\) we rearrange this expression to make \(\frac{dy}{dx}\) the subject, that's: \[2\frac{dy}{dx}y = -6x\] dividing both sides by \(2y\) we obtain: \[\frac{dy}{dx} = -\frac{3x}{y}\]
  2. To find \(\frac{d^2y}{dx^2}\) we differentiate the first differentiated expression we found \[6x+2\frac{dy}{dx}y = 0\] with respect to \(x\) again. Keep in mind that to differentiate the term \(2\frac{dy}{dx}y\) we use the product rule, leading to: \[6+2\frac{d^2y}{dx^2}y+2\frac{dy}{dx}\frac{dy}{dx} = 0\] which can be written: \[6+2\frac{d^2y}{dx^2}y+2\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2 = 0\] We can divide throughout by \(2\) to write: \[3+\frac{d^2y}{dx^2}y+\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2 = 0\] We now rearrange this to make \(\frac{d^2y}{dx^2}\) the subject: \[\frac{d^2y}{dx^2}y = -3 -\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2\] Finally, dividing throughout by \(y\): \[\frac{d^2y}{dx^2} = \frac{-3 -\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2}{y}\] That's: \[\frac{d^2y}{dx^2} = -\frac{3 +\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2}{y}\]

Exercise 2

  1. Given the expression \(sin(y)-cos(x)=1\), find an expression for:
    1. \(\frac{dy}{dx}\)
    2. \(\frac{d^2y}{dx^2}\)
  2. Given the expression \(3x^2y-cos(y)=4\), find an expression for:
    1. \(\frac{dy}{dx}\)
    2. \(\frac{d^2y}{dx^2}\)

Answers Without Working

  1. Given the expression \(sin(y)-cos(x)=1\), we find:
    1. \(\frac{dy}{dx} = -\frac{sin(x)}{cos(y)}\)
    2. \(\frac{d^2y}{dx^2} = \begin{pmatrix}\frac{dy}{dx} \end{pmatrix}^2.tan(y)-\frac{cos(x)}{cos(y)}\)
  2. Given the expression \(3x^2y-cos(y)=4\), we find:
    1. \(\frac{dy}{dx} = -\frac{6xy}{3x^2+sin(y)}\)
    2. \(\frac{d^2y}{dx^2} = - \frac{6y+12x.\frac{dy}{dx} + \begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2.cos(y)}{3x^2+sin(y)}\)