Question

How do you find the derivative of ${{x}^{2}}-{{y}^{2}}=16$?

Answer 1

We have an equation of two variables of x and y. Dependency of the variable is not mentioned. That’s why we find both the differential form and the derivation of y with respect to x from that. The value of $\dfrac{dy}{dx}$ is generally considered as the derivative.

Complete step by step answer:
There are two types of solutions we can get from the derivation of the equation ${{x}^{2}}-{{y}^{2}}=16$.
The one being the derivation of y with respect to x assuming that the y is a dependent function of x.
The other one being the differential form of the equation only.
First, we express the differential form where the base of the differentiation is not a particular variable.
In both cases we apply the differentiation form of $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$.
Now the differential form of that formula will be $d\left( {{x}^{n}} \right)=n{{x}^{n-1}}dx$.
Replacing the value of x with y, we get $d\left( {{y}^{n}} \right)=n{{y}^{n-1}}dy$.
Putting the value of n as $n=2$ in both cases we get
$d\left( {{x}^{2}} \right)=2{{x}^{2-1}}dx=2xdx$ and $d\left( {{y}^{2}} \right)=2{{y}^{2-1}}dy=2ydy$.
The differential of any constant number will be 0.
Now we take differential on both sides of the equation ${{x}^{2}}-{{y}^{2}}=16$ and get
$\begin{aligned} & d\left( {{x}^{2}}-{{y}^{2}} \right)=d\left( 16 \right) \\\ & \Rightarrow d\left( {{x}^{2}} \right)-d\left( {{y}^{2}} \right)=0 \\\ & \Rightarrow 2xdx-2ydy=0 \\\ \end{aligned}$
Therefore, the differential form of the equation ${{x}^{2}}-{{y}^{2}}=16$ is $2xdx-2ydy=0$.
From that differential equation we get the derivation of y with respect to x which is $\dfrac{dy}{dx}$.
$\begin{aligned} & 2xdx-2ydy=0 \\\ & \Rightarrow 2xdx=2ydy \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{2x}{2y}=\dfrac{x}{y} \\\ \end{aligned}$

Therefore, the differentiated form of y with respect to x is $\dfrac{dy}{dx}=\dfrac{x}{y}$.

Note: The opposite of the dependency of the variables can also happen where x is a dependent variable of y which means $x=f\left( y \right)$. In that case the derivative form will be $\dfrac{dx}{dy}$ which will be equal to $\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}$ and the value will be $\dfrac{dx}{dy}=\dfrac{1}{\dfrac{x}{y}}=\dfrac{y}{x}$.