# Parametric Differentiation and Implicit Differentiation

As well as basic differentiation and differentiation using the product, quotient or chain rule there is parametric differentiation and implicit differentiation. In the former, we can apply parametric differentiation when two functions are defined parametrically. In the latter, we can apply implicit differentiation when a function is defined implicitly.

## Parametric Differentiation

Recall that parametric equations may look as follows:

$x=f(t),\hspace{20pt}y=g(t),\hspace{20pt}t_1\leq t\leq t_2$,

where $t$ is the parameter. See more on parametric equations. In order to find the derivative $\frac{dy}{dx}$, we can first find the Cartesian equation or we can use **parametric differentiation**. We can perform parametric differentiation by noting that $\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$ and using the derivatives of the parametric equations:

$\frac{dy}{dx}=\frac{dy}{dt}\div\frac{dx}{dt}=\frac{df}{dt}\div\frac{dg}{dt}$

Hence, we find the derivative of $y$ with respect to $x$ by dividing the derivatives of the original parametric equations. For example, suppose $x=\frac{1}{t}$ and $y=2t$ for $0<t\leq 5$. In this case, $\frac{dx}{dt}=\frac{-1}{t^2}$ and $\frac{dy}{dt}=2$. It follows that $\frac{dy}{dx}=\frac{dy}{dt}\div\frac{dx}{dt}=2\div \frac{-1}{t^2}=-2t^2=-\frac{2}{x^2}$ (since $t=\frac{1}{x}$) . This can also be seen when we find the Cartesian equation first. The Cartesian equation is $y=\frac{2}{x}$ and so $\frac{dy}{dx}=-\frac{2}{x^2}$ as before (more on differentiating polynomials). See Examples 1 and 2 for more parametric differentiation.

## Implicit Differentiation

If a Cartesian equation is defined implicitly, we can use **implicit differentiation** to find an expression for $\frac{dy}{dx}$. In these cases, we must treat $y$ as an unknown function of $x$. For example, consider the implicit equation $xy=4$. We could find the explicit equation $y=\frac{4}{x}$ and so $\frac{dy}{dx}=-\frac{4}{x^2}$ (see more on differentiating polynomials). However, we can also differentiate implicitly by differentiating both sides of $xy=4$ with respect to $x$ as it is. The right hand side becomes 0 but the left hand side is a product (treating $y$ as an unknown function of $x$) and so the product rule should be used:

$\frac{d}{dx}\left(xy\right)=1\times y+x\times \frac{dy}{dx}=y+x\frac{dy}{dx}$

It follows that $y+x\frac{dy}{dx}=0$ which we can rearrange to get $\frac{dy}{dx}=-\frac{y}{x}=-\frac{4}{x^2}$. This is the same expression that we obtained finding the Cartesian equation first. In most cases at this level it is much more complicated to find the explicit equation so using implicit differentiation makes more sense.

Consider finding an expression for $\frac{dy}{dx}$ in a second example of an implicit equation: $x^2y-xy^2=x^3$. Both terms on the left are product whereas the right hand side is a polynomial term. Differentiating both sides with respect to $x$ gives $2xy+x^2\frac{dy}{dx}-y^2-2xy\frac{dy}{dx}=3x^2$. Note that $y^2$ is a chain: $y$ is an unknown function of $x$ that is then squared. Hence, we use the chain rule and the derivative of $y^2$ is $2y\frac{dy}{dx}$. In order ot obtain an expression for the derivative we rearrange to make $\frac{dy}{dx}$ the subject: $\frac{dy}{dx}=\frac{3x^2-2xy+y^2}{x^2-2xy}$. We obtain this by keeping all derivative terms on the left and then factorising. See Examples 3 and 4.

## Examples of Parametric Differentiation and Implicit Differentiation

Find an expression for $\frac{dy}{dx}$ given the parametric equations $x=\frac{t}{t-2}$ and $y=\frac{1}{t+1}$ for $t>2$. Verify your result from the Cartesian equation.

Firstly, we can find an expression for $\frac{dx}{dt}$ using the quotient rule. Let $f(t)=t$ and $g(t)=t-2$. It follows that $f'(t)=1$ and $g'(t)=1$ and so $\frac{dx}{dt}=\frac{1\times (t-2)-t\times 1}{(t-2)^2}=\frac{-2}{(t-2)^2}$. We can write this in terms of $x$ by recalling from Example 1 on the Parametric Equations page that $t=\frac{2x}{x-1}$. Hence, $\frac{dx}{dt}=\frac{-2}{\left(\left(\frac{2x}{x-1}\right)-2\right)^2}=\frac{-2(x-1)^2}{\left(2x-2(x-1)\right)^2}=-\frac{1}{2}(x-1)^2$. Secondly, we can find an expression for $\frac{dy}{dt}$ using the chain rule if we write $y=(t+1)^{-1}$. Let $f(g)=g^{-1}$ and $g(t)=t+1$. It follows that $f'(g)=-g^{-2}$ and $g'(t)=1$ and so $\frac{dy}{dt}=-g^{-2}\times 1=\frac{-1}{(t+1)^2}=\frac{-1}{\left(\left(\frac{2x}{x-1}\right)+1\right)^2}=\frac{-(x-1)^2}{(3x-1)^2}$. Hence, $\frac{dy}{dx}=\frac{dy}{dt}\div \frac{dx}{dt}=\frac{-(x-1)^2}{(3x-1)^2}\div -\frac{1}{2}(x-1)^2=\frac{2}{(3x-1)^2}$.

Recall from Example 1 on the Parametric Equations page that the Cartesian equation is $y=\frac{x-1}{3x-1}$. We can find an expression for $\frac{dy}{dx}$ using the quotient rule. Let $f(x)=x-1$ and $g(x)=3x-1$. It follows that $f'(x)=1$ and $g'(x)=3$ and so $\frac{dy}{dx}=\frac{1\times (3x-1)-(x-1)\times 3}{(3x-1)^2}=\frac{2}{(3x-1)^2}$. This is the same expression we obtained using parametric differentiation. In this case, however, it may have been simpler to find the Cartesian equations first.

Find the equation of the tangent to the curve defined by the parametric equations $x=3\sin(\theta)$ and $y=2\cos(\theta)$ at the points where $\theta=\frac{\pi}{4}$.

The equation of a tangent takes the form $y=mx+c$ and so we must find the $x$ and $y$ coordinates and the gradient at $\theta=\frac{\pi}{4}$. The $x$ and $y$ coordinates are $x=3\sin\left(\frac{\pi}{4}\right)=\frac{3\sqrt{2}}{2}$ and $y=2\cos\left(\frac{\pi}{4}\right)=\sqrt{2}$. To find the gradient, we first find the derivatives of the parametric equations: $\frac{dx}{d\theta}=3\cos(\theta)$ and $\frac{dy}{d\theta}=-2\sin(\theta)$ (see more on differentiating trigonometric functions). It follows that $\frac{dy}{dx}=\frac{dy}{d\theta}\div\frac{dx}{d\theta}=\frac{-2\sin(\theta)}{3\cos(\theta)}=-\frac{2}{3}\tan(\theta)$. At $\theta=\frac{\pi}{4}$, the derivative is $\frac{dy}{dx}=-\frac{2}{3}\tan\left(\frac{\pi}{4}\right)=-\frac{2}{3}$. Since we have the coordinates and gradient we can find the $y$-intercept from $\sqrt{2}=-\frac{2}{3}\times \frac{3\sqrt{2}}{2}+c=-\sqrt{2}+c$. It follows that $c=2\sqrt{2}$ and the equation of the tangent is $y=-\frac{2}{3}x+2\sqrt{2}$.

In this example, it is much simpler to use parametric differentiation than find a Cartesian equation.

Find the value of the gradient of the curve given implicitly by $\frac{3x}{y^2}-\sqrt{xy}=2+(x-y)^3$ at the point $(1,1)$.

Using a combination of the differentiation rules, we can differentiate both sides with respect to $x$ to obtain:

$\begin{array}{l}&&\frac{3}{y^2}-\frac{6x}{y^3}\frac{dy}{dx}-\frac{1}{2}(xy)^{-\frac{1}{2}}\times\left(y+x\frac{dy}{dx}\right)\\&=&3(x-y)^2\times\left(1-\frac{dy}{dx}\right)\end{array}$.

It follows that

$\begin{array}{l}&&\frac{3}{y^2}-\frac{6x}{y^3}\frac{dy}{dx}-\frac{y}{2\sqrt{xy}}-\frac{x}{2\sqrt{xy}}\frac{dy}{dx}\\&=&3(x-y)^2-3(x-y)^2\frac{dy}{dx}\end{array}$

or

$\begin{array}{l}&&3(x-y)^2\frac{dy}{dx}-\frac{6x}{y^3}\frac{dy}{dx}-\frac{x}{2\sqrt{xy}}\frac{dy}{dx}\\&=&3(x-y)^2-\frac{3}{y^2}+\frac{y}{2\sqrt{xy}}\end{array}$.

Hence,

$\frac{dy}{dx}=\frac{3(x-y)^2-\frac{3}{y^2}+\frac{y}{2\sqrt{xy}}}{3(x-y)^2-\frac{6x}{y^3}-\frac{x}{2\sqrt{xy}}}$.

Substituting $x=1$ and $y=1$ we have

$\frac{dy}{dx}=\frac{-3+\frac{1}{2}}{-6-\frac{1}{2}}=\frac{5}{13}$.

Hence, the gradient at the point $(1,1)$ is $\frac{5}{13}$.

Find the exact equation of the normal to the curve defined implicitly by $1-ye^{4x}=3x\ln(2y)$ at the point where $x=0$.

In order to find the equation of the normal we first need the gradient when $x=0$. There are products on both sides of the equation and there is also a chain on the right. So, we use both the product rule and chain rule to differentiate:

$-\frac{dy}{dx}e^{4x}-4ye^{4x}=3\ln(2y)+3x\frac{1}{y}\frac{dy}{dx}$.

It follows that

$\frac{dy}{dx}=-\frac{3\ln(2y)+4ye^{4x}}{e^{4x}+3x\frac{1}{y}}=-\frac{3y\ln(2y)+4y^2e^{4x}}{ye^{4x}+3x}$

Hence, in order to find the gradient at the point where $x=0$, we also require the $y$-coordinate. From the original equation, when $x=0$ it follows that $1-ye^{4\times 0}=3\times 0\times\ln(2y)$ or $1-y=0$. Hence, $y=1$ and so $\frac{dy}{dx}=-\frac{3\ln(2)+4}{1}$. The gradient of the normal is then $m=\frac{1}{3\ln(2)+4}$. We find the $y$-intercept from $1=\frac{1}{3\ln(2)+4}\times 0+c$, hence $c=1$. Hence, the exact equation of the normal at $x=1$ is $y=\frac{1}{3\ln(2)+4}x+1$.