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

Options

A. `(8x^7)/(8^x "log"8)`
B. `(8^x log8)/(8x^7)`
C. `(8^x)/(x^8)[ "log" 8- 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 \(u\)**:
\[ \frac{du}{dx} = 8^x \ln(8) \]
(Using the formula \(\frac{d}{dx} a^x = a^x \ln(a)\))

2. **Differentiate \(v\)**:
\[ \frac{dv}{dx} = 8x^7 \]
(Using the power rule \(\frac{d}{dx} x^n = nx^{n-1}\))

### 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 the values of \(u\), \(v\), \(\frac{du}{dx}\), 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, let's simplify the expression:

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

### Step 5: Factor Out Common Terms
We can factor out \(8^x\) and \(x^7\):

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

This simplifies to:

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

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

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