Derivative of the function f (x) = log5 (Iog,x), x > 7 is

The correct option is 1/x(log5)(log7)(log7−x).

Step-by-Step Explanation:

Let the given function be:
$$f(x)={\log }_5({\log }_{7}x)$$

To find the derivative, we first apply the change of base rule for logarithms, which states that:
$${\log }_{a}b=\frac{{\log }_{e}b}{{\log }_{e}a}=\frac{\log b}{\log a}$$

Using this rule, we can rewrite the function with natural base e as:
$$f(x)=\frac{\log ({\log }_{7}x)}{\log 5}$$

Now, we differentiate both sides with respect to x using the Chain Rule:
$$f^{\prime }(x)=\frac{1}{\log 5}\cdot \frac{d}{dx}[\log ({\log }_{7}x)]$$

Applying the derivative rule for natural logarithm, $\frac{d}{du}(\log u)=\frac{1}{u}$ where $u={\log }_{7}x$:
$$f^{\prime }(x)=\frac{1}{\log 5}\cdot \frac{1}{{\log }_{7}x}\cdot \frac{d}{dx}({\log }_{7}x)$$

Next, we find the derivative of ${\log }_{7}x$ by writing it as $\frac{\log x}{\log 7}$:
$$\frac{d}{dx}({\log }_{7}x)=\frac{1}{\log 7}\cdot \frac{d}{dx}(\log x)=\frac{1}{x\log 7}$$

Substituting this derivative back into our main equation:
$$f^{\prime }(x)=\frac{1}{\log 5}\cdot \frac{1}{{\log }_{7}x}\cdot \frac{1}{x\log 7}$$

Rearranging the terms, we get:
$$f^{\prime }(x)=\frac{1}{x(\log 5)(\log 7)({\log }_{7}x)}$$

Identifying the term ${\log }_{7}x$ in the options (written with a typographic hyphen notation as ${\log }_7-x$), the final expression is matches the correct option:
$$\frac{1}{x(\log 5)(\log 7)(\log 7-x)}$$