2 Differentiation of implicit functions

Fortunately it is not necessary to obtain $y$ in terms of $x$ in order to differentiate a function defined implicitly.

Consider the simple equation

$xy=1$

Here it is clearly possible to obtain $y$ as the subject of this equation and hence obtain $\frac{dy}{dx}$ .

Task!

Express $y$ explicitly in terms of $x$ and find $\frac{dy}{dx}$ for the case $xy=1$ .

Answer

We now show an alternative way of obtaining $\frac{dy}{dx}$ which does not involve writing $y$ explicitly in terms of $x$ at the outset. We simply treat $y$ as an (unspecified) function of $x$ .

Hence if $xy=1$ we obtain

$\frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(1\right).$

The right-hand side differentiates to zero as 1 is a constant. On the left-hand side we must use the product rule of differentiation:

$\frac{d}{dx}\left(xy\right)=x\frac{dy}{dx}+y\frac{dx}{dx}=x\frac{dy}{dx}+y$

Hence $xy=1$ becomes, after differentiation,

$x\frac{dy}{dx}+y=0$ or $\frac{dy}{dx}=-\frac{y}{x}$

In this case we can of course substitute $y=\frac{1}{x}$ to obtain

$y=-\frac{1}{x^2}$

as before.

The method used here is called implicit differentiation and, apart from the final step, it can be applied even if $y$ cannot be expressed explicitly in terms of $x$ . Indeed, on occasions, it is easier to differentiate implicitly even if an explicit expression is possible.

Example 15

Obtain the derivative $\frac{dy}{dx}$ where

$x^2+y=1+y^3$

Solution

We begin by differentiating the left-hand side of the equation with respect to $x$ to get:

$\frac{d}{dx}\left(x^2+y\right)=2x+\frac{dy}{dx}$ .

We now differentiate the right-hand side of with respect to $x$ . Using the chain (or function of a function) rule to deal with the $y^3$ term:

$\frac{d}{dx}\left(1+y^3\right)=\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(y^3\right)=0+3y^2\frac{dy}{dx}$

Now by equating the left-hand side and right-hand side derivatives, we have:

$2x+\frac{dy}{dx}=3y^2\frac{dy}{dx}$

We can make $\frac{dy}{dx}$ the subject of this equation:

$\frac{dy}{dx}-3y^2\frac{dy}{dx}=-2x$ which gives $\frac{dy}{dx}=\frac{2x}{3y^2-1}$

We note that $\frac{dy}{dx}$ has to be expressed in terms of both $x$ and $y$ . This is quite usual if $y$ cannot be obtained explicitly in terms of $x$ . Now try this Task requiring implicit differentiation.

Task!

Find $\frac{dy}{dx}$ if $2y=x^2+siny$

Note that your answer will be in terms of both $y$ and $x$ .

Answer

We sometimes need to obtain the second derivative $\frac{d^{2}y}{d{x}^2}$ for a function defined implicitly.

Example 16

Obtain $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$ at the point $\left(4,2\right)$ on the curve defined by the equation

$x^2-xy-y^2-2y=0$

Solution

Firstly we obtain $\frac{dy}{dx}$ by differentiating the equation implicitly and then evaluate it at $\left(4,2\right)$ .

We have $2x-x\frac{dy}{dx}-y-2y\frac{dy}{dx}-2\frac{dy}{dx}=0$ (1)

from which $\frac{dy}{dx}=\frac{2x-y}{x+2y+2}$ (2)

so at $\left(4,2\right)$ $\frac{dy}{dx}=\frac{6}{10}=\frac{3}{5}$ .

To obtain the second derivative $\frac{d^{2}y}{d{x}^2}$ it is easier to use (1) than (2) because the latter is a quotient. We simplify (1) first:

$2x-y-\left(x+2y+2\right)\frac{dy}{dx}=0$ (3)

We will have to use the product rule to differentiate the third term here.

Hence differentiating (3) with respect to $x$ :

$2-\frac{dy}{dx}-\left(x+2y+2\right)\frac{d^{2}y}{d{x}^2}-\left(1+2\frac{dy}{dx}\right)\frac{dy}{dx}=0$

or

$2-2\frac{dy}{dx}-2{\left(\frac{dy}{dx}\right)}^2-\left(x+2y+2\right)\frac{d^{2}y}{d{x}^2}=0$ (4)

Note carefully that the third term here, ${\left(\frac{dy}{dx}\right)}^2$ , is the square of the first derivative. It should not be confused with the second derivative denoted by $\frac{d^{2}y}{d{x}^2}$ .

Finally, at $\left(4,2\right)$ where $\frac{dy}{dx}=\frac{3}{5}$ we obtain from (4): $2-2\left(\frac{3}{5}\right)-2\left(\frac{9}{25}\right)-\left(4+4+2\right)\frac{d^{2}y}{d{x}^2}=0$

from which $\frac{d^{2}y}{d{x}^2}=\frac{1}{125}$ at $\left(4,2\right)$ .

Task!

This Task involves finding a formula for the curvature of a bent beam. When a horizontal beam is acted on by forces which bend it, then each small segment of the beam will be slightly curved and can be regarded as an arc of a circle. The radius $R$ of that circle is called the radius of curvature of the beam at the point concerned. If the shape of the beam is described by an equation of the form $y=f\left(x\right)$ then there is a formula for the radius of curvature $R$ which involves only the first and second derivatives $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$ .

Find that equation as follows.

Start with the equation of a circle in the simple implicit form

$x^2+y^2=R^2$

and perform implicit differentiation twice. Now use the result of the first implicit differentiation to find a simple expression for the quantity $1+{\left(dy∕dx\right)}^2$ in terms of $R$ and $y$ ; this can then be used to simplify the result of the second differentiation, and will lead to a formula for $\frac{1}{R}$ (called the curvature ) in terms of $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$ .

Answer