Derivative ` 5^(x) w.r.t. log _5 x ` is

Options

A. ` 5^(x) (log 5)`
B. ` x5^(x) (log5)`
C. ` 5^(x) (log 5)^(2)`
D. ` x5^(x) (log 5)^(2)`

Answer

To find the derivative of \( 5^x \) with respect to \( \log_5 x \), we can follow these steps:

### Step 1: Define the Functions
Let:
- \( h = 5^x \)
- \( g = \log_5 x \)

We need to find \( \frac{dh}{dg} \).

### Step 2: Find \( \frac{dh}{dx} \)
Using the formula for the derivative of an exponential function:
\[ \frac{dh}{dx} = \frac{d}{dx}(5^x) = 5^x \log 5 \]

### Step 3: Find \( \frac{dg}{dx} \)
Using the change of base formula for logarithms:
\[ g = \log_5 x = \frac{\log_e x}{\log_e 5} \]
Now, differentiate \( g \):
\[ \frac{dg}{dx} = \frac{1}{\log 5} \cdot \frac{d}{dx}(\log_e x) = \frac{1}{\log 5} \cdot \frac{1}{x} \]
Thus,
\[ \frac{dg}{dx} = \frac{1}{x \log 5} \]

### Step 4: Find \( \frac{dh}{dg} \)
Using the chain rule, we have:
\[ \frac{dh}{dg} = \frac{dh}{dx} \cdot \frac{dx}{dg} = \frac{dh}{dx} \cdot \frac{1}{\frac{dg}{dx}} \]
Substituting the values we found:
\[ \frac{dh}{dg} = \frac{5^x \log 5}{\frac{1}{x \log 5}} = 5^x \log 5 \cdot \frac{x \log 5}{1} \]
This simplifies to:
\[ \frac{dh}{dg} = x \cdot 5^x \cdot (\log 5)^2 \]

### Final Answer
Thus, the derivative \( \frac{d(5^x)}{d(\log_5 x)} \) is:
\[ x \cdot 5^x \cdot (\log 5)^2 \]