Differentiate: `(2^(x))/(x)`

Answer

To differentiate the function \( y = \frac{2^x}{x} \), we will use the quotient rule. The quotient rule states that if you have a function in the form of \( \frac{u}{v} \), then the derivative is given by:

\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

where \( u = 2^x \) and \( v = x \).

### Step 1: Identify \( u \) and \( v \)

Let:
- \( u = 2^x \)
- \( v = x \)

### Step 2: Differentiate \( u \) and \( v \)

Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

- The derivative of \( u = 2^x \) is:
\[ \frac{du}{dx} = 2^x \log(2) \]

- The derivative of \( v = x \) is:
\[ \frac{dv}{dx} = 1 \]

### Step 3: Apply the Quotient Rule

Now we can apply the quotient rule:

\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

Substituting the values we have:

\[ \frac{dy}{dx} = \frac{x \cdot (2^x \log(2)) - (2^x) \cdot (1)}{x^2} \]

### Step 4: Simplify the Expression

Now we simplify the expression:

\[ \frac{dy}{dx} = \frac{x \cdot 2^x \log(2) - 2^x}{x^2} \]

We can factor out \( 2^x \) from the numerator:

\[ \frac{dy}{dx} = \frac{2^x (x \log(2) - 1)}{x^2} \]

### Final Result

Thus, the derivative of \( y = \frac{2^x}{x} \) is:

\[ \frac{dy}{dx} = \frac{2^x (x \log(2) - 1)}{x^2} \]

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