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

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Last updated date: 13th Jun 2024
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Answer
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Hint: We solve the given equation using the identity formula of logarithm where the base of $\ln $ is always $e$. The first step would be to eliminate the logarithm function. Then we first define the multiplication rule and how the differentiation of function works. We take multiplication of these two different differentiated values. We take the $\dfrac{dy}{dx}$ altogether.

Complete step-by-step solution:
We have $\ln a={{\log }_{e}}a$. So, $y=\ln \left( {{x}^{2}}y \right)$ becomes $y={{\log }_{e}}\left( {{x}^{2}}y \right)$.
We know ${{\log }_{e}}a=y\Rightarrow a={{e}^{y}}$. Applying the rule in case of $y={{\log }_{e}}\left( {{x}^{2}}y \right)$, we get
$\begin{align}
  & y={{\log }_{e}}\left( {{x}^{2}}y \right) \\
 & \Rightarrow {{x}^{2}}y={{e}^{y}} \\
\end{align}$
We differentiate the given function ${{x}^{2}}y={{e}^{y}}$ with respect to $x$ using the chain rule.
We now discuss the multiplication process of two functions where \[f\left( x \right)=u\left( x \right)v\left( x \right)\]
Differentiating \[f\left( x \right)=uv\], we get \[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ uv \right]=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\].
The above-mentioned rule is the multiplication rule. We apply that on ${{x}^{2}}y$. We assume the functions where \[u\left( x \right)={{x}^{2}},v\left( x \right)=y\]
We know that differentiation of \[u\left( x \right)={{x}^{2}}\] is ${{u}^{'}}\left( x \right)=2x$ as $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ and differentiation of $v\left( x \right)=y$ is \[{{v}^{'}}\left( x \right)=\dfrac{dy}{dx}\]. We apply the formula of \[\dfrac{d}{dx}\left( {{e}^{y}} \right)={{e}^{y}}\dfrac{dy}{dx}\]. This followed the differential form of chain rule.
We now take differentiation on both parts of ${{x}^{2}}y={{e}^{y}}$ and get \[\dfrac{d}{dx}\left[ {{x}^{2}}y \right]=\dfrac{d}{dx}\left[ {{e}^{y}} \right]\].
We place the chain rule and \[\dfrac{d}{dx}\left( {{e}^{y}} \right)={{e}^{y}}\dfrac{dy}{dx}\] to get \[y\times 2x+{{x}^{2}}\dfrac{dy}{dx}={{e}^{y}}\dfrac{dy}{dx}\].
We take all the $\dfrac{dy}{dx}$ forms altogether to get
\[\begin{align}
  & y\times 2x+{{x}^{2}}\dfrac{dy}{dx}={{e}^{y}}\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}\left( {{e}^{y}}-{{x}^{2}} \right)=2xy \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{2xy}{\left( {{e}^{y}}-{{x}^{2}} \right)} \\
\end{align}\]
We now replace the value of ${{x}^{2}}y={{e}^{y}}$ in the denominator and get
\[\dfrac{dy}{dx}=\dfrac{2xy}{\left( {{x}^{2}}y-{{x}^{2}} \right)}=\dfrac{2xy}{{{x}^{2}}\left( y-1 \right)}=\dfrac{2y}{x\left( y-1 \right)}\].
Therefore, differentiation of $y=\ln \left( {{x}^{2}}y \right)$ is \[\dfrac{2y}{x\left( y-1 \right)}\].

Note: We need to remember that in the chain rule \[\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}\], we aren’t cancelling out the part \[d\left[ h\left( x \right) \right]\]. Cancelation of the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate.