Answer

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**Hint:**First we have to define what the terms we need to solve the problem are. To begin with logarithmic, we will first understand what the logarithmic operator represents in mathematics. Then to find the derivative, we need to first simplify the equation of $y$.We will simplify it using the rules of logarithm and reduce the equation into differential forms. Then we will carry out the differentiation with respect to $x$.

**Complete step by step answer:**

A logarithm function or log operator is used when we have to deal with the powers of a number, to understand it better, we will see an example.

Suppose $y = {z^a}$, where $y,z,a$are the real numbers.

Then if we apply log on both sides with base z, we will get the following results.

$ \Rightarrow {\log _z}y = a$

The low operator has many properties, some of the properties which we will use in this particular problem with differentiation,

The derivative of the logarithmic function is given by: \[f'\left( x \right) = 1/\left( {{\text{ }}x{\text{ }}ln\left( b \right){\text{ }}} \right)x\] is the function argument. b is the logarithm base. \[ln(b)\]is the natural logarithm of b.

Let from the given problem $y = \log x$, now differentiate this log equation with respect to $x$we get

$\dfrac{{dy}}{{dx}} = \dfrac{1}{x}$ (since derivative of $y$will be $\dfrac{{dy}}{{dx}}$and derivative of the $\log x = \dfrac{1}{x}$by the logarithm differentiation)

So, as we find the first derivative of $y = \log x$ is $\dfrac{{dy}}{{dx}} = \dfrac{1}{x}$

Now take $\dfrac{{dy}}{{dx}} = {y_1}$(first derivative function) so then ${y_1} = \dfrac{1}{x}$ and cross multiply we get $x{y_1} = 1$

Now again in equation $x{y_1} = 1$ differentiation with respect to $x$ we get

$x{y_2} + {y_1} = 0$(in left hand we are applying UV derivative and in right hand side derivative of $1$ is $0$ and ${y_2}$ represent the double derivative of y)

Hence from this $y = \log x$ is a solution of $x{y_2} + {y_1} = 0$

**Note:**To apply the UV method of differentiation, first choose U and V, differentiate U and integrate V

Thus \[u\smallint vdx - \smallint u'\left( {\smallint vdx} \right)dx.\]and differentiation of zero will be constant.

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