11. Parametric and Implicit Differentiation

11.1. Parametric Equations

The idea behind parametric equations is that we can describe a curve in the \((x,y)\)-plane by giving its \(x\)- and \(y\)-coordinates in terms of a third variable, known as a parameter. Often this parameter will be \(t\) (which we can interpret as time). In this case we can think of the curve being formed as time increases.

For example, we could have the equations \(x = 2t+1\) and \(y = t^2\) for all \(t \in \mathbb{R}\). Then at time \(t\) we get the point on the curve \((2t+1,t^2)\), e.g. when \(t=0\) we get the point \((1,0)\) and when \(t=1\) we get the point \((3,1)\). If we plot some points on this curve for various \(t\) we can start to get an idea of what the curve looks like. But is there a better way?

Indeed there is. We can eliminate \(t\) from our equations to express \(y\) in terms of \(x\). Since \(x=2t+1\) we rearrange to get

\[ t = \frac{x-1}{2}. \]

Substituting this into the equation for \(y\) we get

\[ y = \left(\frac{x-1}{2}\right)^2 = \frac{(x-1)^2}{4}. \]

So we have two ways of expressing the same curve. Here, \(x = 2t+1\), \(y = t^2\) is known as the parametric equation for the curve and \(y = \frac{(x-1)^2}4\) is known as the cartesian equation.

Fig. 11.1 Graph of \(y=\frac{(x-1)^2}{4}\).

In the parametric equation, \(t\) is called the parameter.

11.2. Examples

Example 11.1

Let \(x = t\), \(y = 2t\) for \(t \in \mathbb{R}\). What is the cartesian equation of this curve?

Solution.
We start to consider some values for \(t\) to get an idea.

When \(t = 0\) we get the point \((0,0)\). When \(t = 1\) we get \((1,2)\), and when \(t = -1\) we get \((-1,-2)\). This is looking like a straight line!

We eliminate \(t\) to confirm our guess. We have \(t = x\) so \(y = 2t = 2x\) which is a straight line, as we’d hoped.

Thus the cartesian equation of the curve is the straight line \(y=2x\).

Example 11.2

Let \(x = t-1\), \(y = 2t+1\) for \(t \in \mathbb{R}\). What is the cartesian equation of this curve?

Solution.
We start to consider some values for \(t\) to get an idea. If \(t = 0\) we get \((-1,1)\). If \(t = 1\) we get \((0,3)\) and if \(t = -1\) we get \((-2,1)\). Again, this is looking like a straight line.

We eliminate \(t\) to find the cartesian equation. We have \(x= t-1\) so \(t = x+1\) and hence

\[ y = 2(x+1)+1 = 2x + 3. \]

Thus the cartesian equation of the curve is \(y=2x+3\).

11.3. Parametric Differentiation

If we start out with a parametric equation for a curve, we might want to find the gradient of the curve at a point, that is find \(\frac{dy}{dx}\). How do we go about doing this?

Example 11.3

Let \(x = t^2+t\), \(y = t^3+1\) for \(t \in \mathbb{R}\). Find \(\frac{dy}{dx}\).

Solution.
If we could find a nice form for the cartesian equation of the curve then we could calculate \(\frac{dy}{dx}\) from that, but in fact it’s not looking too easy. We’ll find another way.

We know from the chain rule that \(\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}\). But, from the inverse function rule \(\frac{dx}{dt}=\frac{1}{\frac{dt}{dx}}\) so that \(\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}\). Hence we find that

\[ \frac{dy}{dx}=\frac{~\frac{dy}{dt}~}{\frac{dx}{dt}}. \]

Now \(\frac{dx}{dt} = 2t+1\) and \(\frac{dy}{dt} = 3t^2\) so that

\[ \frac{dy}{dx} = \frac{3t^2}{2t+1} \mbox{ for } t \neq -\frac 12. \]

Hence we have found \(\frac{dy}{dx}\) in terms of \(t\). Since we started with both \(x\) and \(y\) in terms of \(t\), this is fine.

We summarise our method in the following theorem.

Theorem 11.1 (Parametric Differentiation)

Let \(x = f(t)\) and \(y = g(t)\) be a parametric equation. Then

\[ \frac{dy}{dx} = \frac{~\frac{dy}{dt}~}{\frac{dx}{dt}}. \]

Example 11.4

Consider the example of Section 11.1 that is, \(x = 2t +1\) and \(y = t^2\) for all \(t \in \mathbb{R}\). Find \(\frac{dy}{dx}\) in terms of \(t\) and in terms of \(x\).

Solution.
We have \(\frac{dx}{dt} = 2\) and \(\frac{dy}{dt} = 2t\) so that \(\frac{dy}{dx} = \frac{~\frac{dy}{dt}~}{\frac{dx}{dt}} = \frac{2t}{2} = t\) for all \(t \in \mathbb{R}\).

To find \(\frac{dy}{dx}\) in terms of \(x\), we could now use the fact that \(x = 2t +1\) and thus \(t = \frac{x-1}{2}\) which means that \(\frac{dy}{dx} = t = \frac{x-1}{2}\). Or we can use the cartesian equation we worked out before as \(y = \frac{(x-1)^2}{4}\). Here we can differentiate \(y\) by using the chain rule and get \(\frac{dy}{dx} = \frac 14 . 2(x-1) = \frac {x-1}2\) as expected.

11.4. Implicit Equations

Consider the equation

\[ yx-1 = 3x+2y. \]

Here \(y\) is related to \(x\) even though it is not written in the form \(y = f(x)\) for some function \(f(x)\). Such an equation is said to be an implicit equation.

Sometimes we can rearrange such an equation to make \(y\) the subject. Indeed, in the above case we get

\[\begin{align*} yx - 1 =3x+2y & \mbox{ so } &yx - 2y = 3x+1\\ & \mbox{ so } & y(x-2) = 3x+1\\ & \mbox{ so } & y =\frac{3x+1}{x-2}\mbox{ for } x \neq 2. \end{align*}\]

This known as the explicit form of the equation.

11.5. Implicit Differentiation

We would like to develop a way to find \(\frac{dy}{dx}\) starting with an implicit equation.

Example 11.5

Let \(y^3 + y = x^2 +x\). We will try to find \(\frac{dy}{dx}\) by differentiating both sides with respect to \(x\).

Differentiating the right-hand side with respect to \(x\) is easy, but the left-hand side looks more problematic. We will take one term at a time.

Let \(u=y^3\). We want to differentiate \(u\) with respect to \(x\), that is find \(\frac{du}{dx}\). Since \(u\) is written in terms of \(y\) we use the chain rule to get

\[ \frac{du}{dx} = \frac{du}{dy}\frac{dy}{dx}. \]

But \(\frac{du}{dy}=3y^2\), so we get \(\frac{du}{dx}=3y^2\frac{dy}{dx}\), in other words

\[ \frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}. \]

We can do the same trick again to find the derivative of of the term ‘\(y\)’ with respect to~\(x\): putting \(w=y\) we get

\[ \frac{dw}{dx} = \frac{dw}{dy}\frac{dy}{dx} = 1.\frac{dy}{dx} = \frac{dy}{dx} \]

so that \(\frac{d}{dx}(y)=\frac{dy}{dx}\) (perhaps not so surprising!). Hence differentiating our equation \(y^3+y=x^2+x\) with respect to \(x\) gives us

\[ 3y^2\frac{dy}{dx} + \frac{dy}{dx} = 2x + 1 \]

and we can rearrange to get

\[ \frac{dy}{dx}(3y^2+1)=2x+1 \]

and hence

\[ \frac{dy}{dx} = \frac{2x+1}{3y^2+1}. \]

(Note that since \(3y^2+1> 0\) for all \(y\), we have no problems dividing here.)\

Remark 11.1

Since the equation we started with relating \(y\) and \(x\) was an implicit equation, we don’t have a problem with \(\frac{dy}{dx}\) being written in terms of both \(x\) and \(y\).

Generalising the methods used in the above example we get the following.

Theorem 11.2 (Differentiating \(f(y)\) with respect to \(x\))

\[ \frac{d}{dx}(f(y))=f'(y).\frac{dy}{dx} \]

(That is, to differentiate a function of \(y\) with respect to \(x\), first differentiate with respect to \(y\) then multiply by \(\frac{dy}{dx}\).)

Example 11.6

Let \(y^3 + 4x = x^2 + \frac{1}{y}\) for \(y\neq 0\). Find \(\frac{dy}{dx}\).

Solution.
As before, we start with our implicit equation \(y^3 + 4x = x^2 + \frac{1}{y}\) and differentiate both sides with respect to \(x\).

By Theorem 11.2, to differentiate the terms which are a function of \(y\) with respect to \(x\), we first differentiate with respect to \(y\) and then multiply by \(\frac{dy}{dx}\). Hence, for \(y\neq 0\) we get

\[ 3y^2 \frac{dy}{dx} + 4 = 2x -\frac{1}{y^2}\frac{dy}{dx}. \]

We now solve this equation for \(\frac{dy}{dx}\) to get

\[\begin{align*} 3y^2 \frac{dy}{dx} + 4 = 2x -\frac{1}{y^2}\frac{dy}{dx} & \mbox{ so } & 3y^4 \frac{dy}{dx} + 4y^2 = 2xy^2 - \frac{dy}{dx}\\ & \mbox{ so } & 3y^4 \frac{dy}{dx} + \frac{dy}{dx} = 2xy^2-4y^2\\ & \mbox{ so } & \frac{dy}{dx}(3y^4 + 1) = y^2(2x-4)\\ & \mbox{ so } & \frac{dy}{dx} = \frac{y^2(2x-4)}{3y^4 + 1}\quad\mbox{since $3y^4 + 1> 0$ for all $y$.} \end{align*}\]

Thus, for \(y\neq 0\), we get \(\displaystyle \frac{dy}{dx} = \frac{y^2(2x-4)}{3y^4 + 1}\). (Note that if \(y = 0\), then the function is as such not defined and so there can’t be a derivative of it at \(y = 0\).)

Example 11.7

A curve has implicit equation \(x^{-2} + x^3 y^5 + y^2 = \frac{37}{4}\).

(a) Find \(\frac{dy}{dx}\).

(b) Does the point \((3,4)\) lie on the curve? If so, find the gradient of the curve at \((3,4)\).

(c) Does the point \((2,1)\) lie on the curve? If so, find the gradient of the curve at \((2,1)\).

Solution.
(a) Working term by term we differentiate both sides of the equation with respect to \(x\) to get

\[ -2x^{-3} +\frac{d}{dx}(x^3y^5) + 2y\frac{dy}{dx} = 0. \]

Note that to differentiate the term \(x^3 y^5\) (which is a product of a function of \(x\) and a function of \(y\)) we need the product rule:

\[\begin{align*} \frac{d}{dx}(x^3 y^5) &= x^3\frac{d}{dx}(y^5) + y^5 \frac{d}{dx}(x^3)\\ &= x^3(5y^4)\frac{dy}{dx} + y^5(3x^2). \end{align*}\]

Hence we have

\[ -2x^{-3} +5x^3 y^4\frac{dy}{dx} + 3x^2 y^5 + 2y\frac{dy}{dx} = 0 \]

so that

\[ \frac{dy}{dx}(5x^3y^4+2y) = 2x^{-3} - 3x^2y^5 \]

and hence

\[ \frac{dy}{dx} = \frac{2x^{-3} - 3x^2y^5}{5x^3y^4 +2y} \mbox{ for }5x^3y^4 +2y \neq 0. \]

(b) If we put \(x = 3\) and \(y = 4\) into the equation for the curve we get

\[ \mbox{LHS} = 3^{-2} + 3^3 4^5+4^2 = \frac{1}{9} + 27\times 1024 + 16 \neq \frac{37}{4} = \mbox{RHS}. \]

Hence the point \((3,4)\) does not lie on the curve.

(c) We put \(x = 2\) and \(y =1\) into the equation to get

\[ \mbox{LHS} = 2^{-2} + 2^3 1^5+1^2 = \frac 14 +8 + 1 = 9+\frac 14 = \frac{37}{4} = \mbox{RHS}. \]

Thus the point \((2,1)\) does lie on the curve.

To find the gradient at \((2,1)\) we put \(x =2\) and \(y=1\) in our expression for \(\frac{dy}{dx}\) to get

\[ \frac{dy}{dx} = \frac{(2\times 2^{-3}) - (3\times 2^2\times 1^5)}{(5\times 2^3\times 1^5)+(2\times1)} = \frac{\frac 14 -12}{40+2} = \frac{-\frac{47}{4}}{42} = - \frac{47}{168}. \]

Thus the gradient at \((2,1)\) is \(-\frac{47}{168}\).

11.6. Taking Logarithms to help with Differentiation

Recall that if \(y = \ln(f(x))\) then \(\displaystyle \frac{dy}{dx} = \frac{f'(x)}{f(x)}\). We can use this, together with implicit differentiation, to sometimes simplify our work when dealing with certain types of functions.

Example 11.8

Let \(\displaystyle y = \frac{x(x^3+1)}{(x^2+3)^2}\) and find \(\displaystyle \frac{dy}{dx}\).

Solution.
We begin by taking natural logarithms to get

\[\begin{align*} \ln y &= \displaystyle \ln \left(\frac{x(x^3+1)}{(x^2+3)^2}\right)\\ &= \ln(x(x^3+1)) - \ln((x^2+3)^2)\\ &= \ln x + \ln(x^3+1) - 2\ln(x^2+3). \end{align*}\]

Differentiating this using implicit differentiation gives

\[ \frac{1}{y}\frac{dy}{dx} = \frac 1x + \frac{3x^2}{x^3+1} -\frac{4x}{x^2+3}. \]

Thus

\[\begin{align*} \frac{dy}{dx} &= y\left(\frac 1x + \frac{3x^2}{x^3+1} -\frac{4x}{x^2+3}\right)\\ &= \frac{x(x^3+1)}{(x^2+3)^2}\left(\frac 1x + \frac{3x^2}{x^3+1} -\frac{4x}{x^2+3}\right)\\ &= \frac{x^3+1}{(x^2+3)^2}+ \frac{3x^3}{(x^2+3)^2} - \frac{4x^2(x^3+1)}{(x^2+3)^3}\\ &= \frac{(x^3+1)(x^2+3)+3x^3(x^2+3)-4x^2(x^3+1)}{(x^2+3)^3}\\ &= \frac{(x^5+3x^3+x^2+3) + (3x^5+9x^3)-(4x^5+4x^2)}{(x^2+3)^3}\\ &= \frac{12x^3-3x^2+3}{(x^2+3)^3}\\ &= \frac{3(4x^3-x^2+1)}{(x^2+3)^3}. \end{align*}\]

As a challenge, try it without using the rules of logarithms first.

Another application of implicit differentiation is when the variable is in an exponent.

Example 11.9

Let \(y = 3^x\) and find \(\frac{dy}{dx}\).

Solution.
Take natural logarithms to get

\[ \ln y = \ln (3^x) = x\ln 3. \]

Note that \(\ln 3\) is just a number (approximately \(1.099\)). We now use implicit differentiation to get

\[ \frac{1}{y}\frac{dy}{dx} = \ln 3\]

so that

\[ \frac{dy}{dx} = y \ln 3 = 3^x\ln 3. \]

Example 11.10

Let \(y = x^x\) for \(x > 0\). Find \(\frac{dy}{dx}\).

Solution.
We take natural logarithms to get

\[ \ln y = \ln x^x = x\ln x. \]

If we want to differentiate the term \(x\ln x\) we need the product rule with \(u = x\) and \(v = \ln x\). We then get

\[ \frac{d}{dx}(x\ln x)=v\frac{du}{dx} + u\frac{dv}{dx} = (\ln x).1 + x.\frac{1}{x}=\ln x+1. \]

Hence, differentiating the equation \(\ln y=x\ln x\), we get

\[ \frac 1y \frac{dy}{dx} = \ln x+1\]

so that

\[ \frac{dy}{dx} = y(\ln x+1) = x^x(\ln x+1). \]