Question

For constant a, \(\frac{d}{dx}\)(x^x + x^a+ a^x + a^a) is

• A. x^x(1 + logx) + ax^a-1
• B. x^x(1 + logx) + ax^a-1 + a^x log a
• C. x^x(1 + logx) + a^a (1 + logx)
• D. x^x(1 + logx) + a^a (1 + loga) + ax^a-1

Answer 1

The Correct Option is B

We are tasked with differentiating the expression: \[ f(x) = x^x + x^a + a^x + a^a \] We will differentiate each term separately. 1. Differentiating \( x^x \): The derivative of \( x^x \) can be computed using logarithmic differentiation. Take the natural logarithm of \( x^x \): \[ \ln(y) = \ln(x^x) = x \ln(x) \] Now differentiate both sides with respect to \( x \): \[ \frac{d}{dx} \left( \ln(y) \right) = \frac{d}{dx} \left( x \ln(x) \right) \] Using the product rule: \[ \frac{1}{y} \frac{dy}{dx} = \ln(x) + 1 \] So: \[ \frac{dy}{dx} = x^x (1 + \log x) \] 2. Differentiating \( x^a \): The derivative of \( x^a \) (where \( a \) is constant) is: \[ \frac{d}{dx} \left( x^a \right) = a x^{a-1} \] 3. Differentiating \( a^x \): The derivative of \( a^x \) (where \( a \) is constant) is: \[ \frac{d}{dx} \left( a^x \right) = a^x \log a \] 4. **Differentiating \( a^a \): Since \( a^a \) is a constant, its derivative is 0. Now, summing all the derivatives: \[ \frac{d}{dx} \left( x^x + x^a + a^x + a^a \right) = x^x (1 + \log x) + a x^{a-1} + a^x \log a \]

Therefore, the correct answer is (B).

Answer 2

Let \( y = x^x + x^a + a^x + a^a \). We want to find \( \frac{dy}{dx} \).

We can differentiate each term separately:

1. For \( x^x \), let \( z = x^x \). Then \( \ln z = x \ln x \). Differentiating both sides with respect to \( x \), we get:

\( \frac{1}{z} \frac{dz}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1 \)

\( \frac{dz}{dx} = z(\ln x + 1) = x^x(1 + \ln x) \)

2. For \( x^a \), \( \frac{d}{dx}(x^a) = ax^{a-1} \)

3. For \( a^x \), \( \frac{d}{dx}(a^x) = a^x \ln a \)

4. For \( a^a \), since \( a \) is a constant, \( a^a \) is also a constant, so \( \frac{d}{dx}(a^a) = 0 \)

Therefore, \( \frac{dy}{dx} = x^x(1 + \ln x) + ax^{a-1} + a^x \ln a + 0 \)

\( \frac{dy}{dx} = x^x(1 + \ln x) + ax^{a-1} + a^x \ln a \)