Differentiation of log₅ (log x²) w.r.t.x is

The correct answer is:
$$\frac{2}{x\left(\log 5\right)\left(\log x\right)}$$

Step-by-step Derivation:

Let the given function be:
$$y={\log }_5\left(\log x^2\right)$$

First, we can simplify the inner term using the logarithm power rule, which states that $\log a^b=b\log a$:
$$\log x^2=2\log x$$
Substituting this back into our function, we get:
$$y={\log }_5\left(2\log x\right)$$

Next, we use the base-change formula for logarithms, ${\log }_{a}b=\frac{\log b}{\log a}$ (where the logarithms on the right-hand side are natural logarithms, base e):
$$y=\frac{\log \left(2\log x\right)}{\log 5}$$

Now, we differentiate $y$ with respect to $x$. Since $\frac{1}{\log 5}$ is a constant, we can pull it out:
$$\frac{dy}{dx}=\frac{1}{\log 5}·\frac{d}{dx}\left[\log \right(2\log x\left)\right]$$

Applying the chain rule, $\frac{d}{dx}\left[\log u\right]=\frac{1}{u}·\frac{du}{dx}$ where $u=2\log x$:
$$\frac{dy}{dx}=\frac{1}{\log 5}·\frac{1}{2\log x}·\frac{d}{dx}\left(2\log x\right)$$

Since the derivative of $2\log x$ is $\frac{2}{x}$:
$$\frac{dy}{dx}=\frac{1}{\log 5}·\frac{1}{2\log x}·\frac{2}{x}$$

Simplifying the expression by canceling the factor of 2 in the numerator and denominator:
$$\frac{dy}{dx}=\frac{2}{x\left(\log 5\right)\left(\log x\right)}$$