Question

How do you find the derivative of $y=\ln \left( {{x}^{2}}+{{y}^{2}} \right)$?

Answer 1

Hint: We will differentiate the given expression using implicit differentiation. We will look at the concept of implicit differentiation. We will use the chain rule, which is given as $\dfrac{d\left( f\circ g\left( x \right) \right)}{dx}={f}'\left( g\left( x \right) \right)\cdot {g}'\left( x \right)$. We will also use the rule for differentiation of addition of functions. Then we will obtain an expression that has the term $\dfrac{dy}{dx}$ on both sides. We will rearrange the equation to obtain the required derivative.

Complete step by step answer:
The given equation is $y=\ln \left( {{x}^{2}}+{{y}^{2}} \right)$. We will differentiate this function using the concept of implicit differentiation. In implicit differentiation, we consider one variable to be dependent on the other. In the given equation, we have two variables $x$ and $y$. We can see from the given equation that the variable $y$ is dependent on the variable $x$. So, upon using implicit differentiation, we have
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \ln \left( {{x}^{2}}+{{y}^{2}} \right) \right)$
We can see that the right hand side of the given equation is a composition of two functions. We will use the chain rule to differentiate this composition of functions. We know that the chain rule is given as $\dfrac{d\left( f\circ g\left( x \right) \right)}{dx}={f}'\left( g\left( x \right) \right)\cdot {g}'\left( x \right)$. Using this rule, we get the following
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \ln \left( {{x}^{2}}+{{y}^{2}} \right) \right)\times \dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}} \right)$
We know that the derivative of the logarithm function is $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$. Therefore, we get
$\dfrac{dy}{dx}=\dfrac{1}{{{x}^{2}}+{{y}^{2}}}\times \dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}} \right)$
Differentiation is a linear function. Therefore, we have the rule for differentiation of sum of functions which is $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)$. Using this rule we get,
$\dfrac{dy}{dx}=\dfrac{1}{{{x}^{2}}+{{y}^{2}}}\times \left( \dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right) \right)$
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}$. Therefore, we have
$\dfrac{dy}{dx}=\dfrac{1}{{{x}^{2}}+{{y}^{2}}}\times \left( 2x+2y\dfrac{dy}{dx} \right)$
Let $\dfrac{dy}{dx}={y}'$. So, substituting this in the above equation, we get
${y}'=\dfrac{2x+2y{y}'}{{{x}^{2}}+{{y}^{2}}}$
Rearranging and solving for ${y}'$, we get
$\begin{align}
  & {y}'\left( {{x}^{2}}+{{y}^{2}} \right)=2x+2y{y}' \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{2x}{{{y}'}}+2y \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-2y=\dfrac{2x}{{{y}'}} \\
 & \therefore {y}'=\dfrac{2x}{{{x}^{2}}+{{y}^{2}}-2y} \\
\end{align}$

So, the differentiation we obtain is $\dfrac{dy}{dx}=\dfrac{2x}{{{x}^{2}}+{{y}^{2}}-2y}$.

Note: It is important to know the rules of differentiation for such types of questions. We should be familiar with the derivatives of the standard functions as they are useful in calculating derivatives of complicated functions. It is better to write every step while doing implicit differentiation since it allows us to keep a check on the dependent and the independent variable.