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How to use logarithmic differentiation to find the derivative of a function, and can it be used for any function?

Answer
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Hint: In this question, we have to use the logarithm differentiation for a function. As we know, the logarithm function is an exponential function. So, to solve this problem, we will let a function $y=f(x)$ and put a log function on both sides of the equation. After that, we will differentiate the log equation with respect to x. then, we will multiply y on both sides of the equation. In the last, we will put the value of the given equation in the new equation and make the necessary calculations, which is the required solution to the problem.

Complete step-by-step answer:
According to the question, we have to find the derivative of a function using logarithmic differentiation.
Thus, we will let a function $y=f(x)$ ------ (1)
Now, we will put log function on both sides in the equation (1), we get
$\log (y)=\log \left( f(x) \right)$
So, now we will differentiate the above equation with respect to x, we get
$\dfrac{dy}{dx}.\dfrac{1}{y}=\dfrac{1}{f(x)}.{f}'(x)$
Now, we will multiply y on both sides in the above equation, we get
$\dfrac{dy}{dx}.\dfrac{1}{y}.y=\dfrac{1}{f(x)}.{f}'(x).y$
On further solving, we get
$\dfrac{dy}{dx}=\dfrac{y{f}'(x)}{f(x)}$
Thus, we will put the value of equation (1) in the above equation, we get
$\dfrac{dy}{dx}=\dfrac{f(x){f}'(x)}{f(x)}$
On further simplification, we get
$\dfrac{dy}{dx}={f}'(x)$
Therefore, we find the differentiation of a function using the logarithm function. Also, we get to know that we can use logarithm differentiation in any function.

Note: While solving this problem, do mention all the steps properly to avoid confusion and mathematical error. Always, remember the best way to use log derivative when you have either an exponential function or a variable exponent function. Since we have to find the derivative of a function, so if you take an example like $y=x,$ $y={{x}^{x}},$ or $y={{x}^{2}}-1$ and use logarithmic differentiation, you will get the accurate answer.