Evaluate `dy/dx`, if `y=8^x/x^8`.

Options

A. `(8x^7)/(8^xlog8)`
B. `(8^xlog8)/(8x^7)`
C. `(8^x)/(x^8)[log8-8/x]`
D. None of these

Answer

To evaluate \(\frac{dy}{dx}\) for the function \(y = \frac{8^x}{x^8}\), we will use the quotient rule of differentiation. The quotient rule states that if \(y = \frac{u}{v}\), then:

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

where \(u = 8^x\) and \(v = x^8\).

### Step 1: Identify \(u\) and \(v\)
Let:
- \(u = 8^x\)
- \(v = x^8\)

### Step 2: Differentiate \(u\) and \(v\)
Now we need to find \(\frac{du}{dx}\) and \(\frac{dv}{dx}\).

1. Differentiate \(u\):
\[ \frac{du}{dx} = \frac{d}{dx}(8^x) = 8^x \ln(8) \]

2. Differentiate \(v\):
\[ \frac{dv}{dx} = \frac{d}{dx}(x^8) = 8x^7 \]

### Step 3: Apply the Quotient Rule
Now we can apply the quotient rule:

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

Substituting \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\):

\[ \frac{dy}{dx} = \frac{x^8 (8^x \ln(8)) - 8^x (8x^7)}{(x^8)^2} \]

### Step 4: Simplify the Expression
Now simplify the expression:

\[ \frac{dy}{dx} = \frac{8^x x^8 \ln(8) - 8^{x+1} x^7}{x^{16}} \]

Factor out \(8^x x^7\):

\[ \frac{dy}{dx} = \frac{8^x x^7 (x \ln(8) - 8)}{x^{16}} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{8^x (x \ln(8) - 8)}{x^9} \]

### Final Result
Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{8^x (x \ln(8) - 8)}{x^9} \]