Parametric Differentiation and Implicit Differentiation

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.

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.

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 $\frac{5}{13}$.

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 $y=\frac{1}{3\ln(2)+4}x+1$.