What is the derivative of `log_(x)5` with respect to `log_(5)x`?

Options

A. `-(log_(5)x)^(-2)`
B. `(log_(5)x)^(-2)`
C. `-(log_(x)5)^(-2)`
D. `(log_(x)5)^(-2)`

Explanation

To find the derivative of \( \log_{x} 5 \) with respect to \( \log_{5} x \), we can follow these steps:

### Step 1: Change of Base Formula
We can express both logarithms using the change of base formula:
\[ \log_{x} 5 = \frac{\log_{e} 5}{\log_{e} x} \]
\[ \log_{5} x = \frac{\log_{e} x}{\log_{e} 5} \]

### Step 2: Let \( u \) and \( v \)
Let:
\[ u = \log_{x} 5 = \frac{\log_{e} 5}{\log_{e} x} \]
\[ v = \log_{5} x = \frac{\log_{e} x}{\log_{e} 5} \]

### Step 3: Differentiate \( u \) with respect to \( x \)
To find \( \frac{du}{dx} \), we will use the quotient rule:
\[ \frac{du}{dx} = \frac{(0 \cdot \log_{e} x) - (\log_{e} 5 \cdot \frac{d}{dx}(\log_{e} x))}{(\log_{e} x)^{2}} \]
Since \( \frac{d}{dx}(\log_{e} x) = \frac{1}{x} \), we have:
\[ \frac{du}{dx} = \frac{0 - \log_{e} 5 \cdot \frac{1}{x}}{(\log_{e} x)^{2}} = -\frac{\log_{e} 5}{x (\log_{e} x)^{2}} \]

### Step 4: Differentiate \( v \) with respect to \( x \)
Now, we differentiate \( v \):
\[ \frac{dv}{dx} = \frac{(\log_{e} 5 \cdot \frac{d}{dx}(\log_{e} x)) - (\log_{e} x \cdot 0)}{(\log_{e} 5)^{2}} \]
This simplifies to:
\[ \frac{dv}{dx} = \frac{\log_{e} 5 \cdot \frac{1}{x}}{(\log_{e} 5)^{2}} = \frac{1}{x \log_{e} 5} \]

### Step 5: Find \( \frac{du}{dv} \)
Now we can find the derivative of \( u \) with respect to \( v \):
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{-\frac{\log_{e} 5}{x (\log_{e} x)^{2}}}{\frac{1}{x \log_{e} 5}} \]
This simplifies to:
\[ \frac{du}{dv} = -\frac{\log_{e} 5}{(\log_{e} x)^{2}} \cdot \frac{x \log_{e} 5}{1} = -\frac{(\log_{e} 5)^{2}}{(\log_{e} x)^{2}} \]

### Final Result
Thus, the derivative of \( \log_{x} 5 \) with respect to \( \log_{5} x \) is:
\[ \frac{du}{dv} = -\frac{(\log_{e} 5)^{2}}{(\log_{e} x)^{2}} \]

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