To find the first derivative of the given expression, we need to apply the rules of differentiation. The expression is initially presented in a form that can be simplified using fundamental properties of logarithms and exponentials. After simplifying, we will differentiate each term separately and then combine the results.

Expression Simplification

The given expression is \(e^{x \ln a} + e^{a \ln x} + e^{a \ln a}\).

We can simplify each term using two key logarithm and exponential properties:

• The property \(n \ln m = \ln (m^n)\), which allows us to move a coefficient inside the logarithm as a power.
• The property \(e^{\ln P} = P\), which states that the exponential function and the natural logarithm are inverse operations, effectively canceling each other out.

• Simplifying the first term, \(e^{x \ln a}\):

  Using the property \(n \ln m = \ln (m^n)\), we rewrite \(x \ln a\) as \(\ln (a^x)\). So, the term becomes \(e^{\ln (a^x)}\). Applying the property \(e^{\ln P} = P\), we simplify \(e^{\ln (a^x)}\) to \(a^x\).

• Simplifying the second term, \(e^{a \ln x}\):

  Similarly, using \(n \ln m = \ln (m^n)\), we rewrite \(a \ln x\) as \(\ln (x^a)\). So, the term becomes \(e^{\ln (x^a)}\). Applying \(e^{\ln P} = P\), we simplify \(e^{\ln (x^a)}\) to \(x^a\).

• Simplifying the third term, \(e^{a \ln a}\):

  Using \(n \ln m = \ln (m^n)\), we rewrite \(a \ln a\) as \(\ln (a^a)\). So, the term becomes \(e^{\ln (a^a)}\). Applying \(e^{\ln P} = P\), we simplify \(e^{\ln (a^a)}\) to \(a^a\).

Therefore, the simplified expression, let's denote it as \(y\), is:

\[y = a^x + x^a + a^a\]

Derivative Calculation

Now, we need to find the first derivative of \(y\) with respect to \(x\), which is \(\frac{dy}{dx}\). We will differentiate each simplified term separately:

• Derivative of \(a^x\):

  This is an exponential function where the base \(a\) is a constant and the exponent \(x\) is the variable. The standard derivative formula for this type of function is \(\frac{d}{dx}(b^x) = b^x \ln b\).

  So, for \(a^x\), its derivative is \(\frac{d}{dx}(a^x) = a^x \ln a\).

• Derivative of \(x^a\):

  This is a power function where the base \(x\) is the variable and the exponent \(a\) is a constant. We use the power rule for differentiation, which states \(\frac{d}{dx}(x^n) = n x^{n-1}\).

  So, for \(x^a\), its derivative is \(\frac{d}{dx}(x^a) = a x^{a-1}\).

• Derivative of \(a^a\):

  In this term, both the base \(a\) and the exponent \(a\) are constants with respect to \(x\). The derivative of any constant value is \(0\).

  So, for \(a^a\), its derivative is \(\frac{d}{dx}(a^a) = 0\).

Derivative Combination

To find the first derivative of the entire expression, we sum the derivatives of its individual terms:

\[\frac{dy}{dx} = \frac{d}{dx}(a^x) + \frac{d}{dx}(x^a) + \frac{d}{dx}(a^a)\]

Substituting the derivatives we found:

\[\frac{dy}{dx} = a^x \ln a + a x^{a-1} + 0\]

Thus, the first derivative is:

\[\frac{dy}{dx} = a^x \ln a + a x^{a-1}\]

Final Derivative Result

The first derivative of the given expression \(e^{x \ln a} + e^{a \ln x} + e^{a \ln a}\) is \(a^x \ln a + a x^{a-1}\).

Comparing this result with the provided options:

The calculated derivative matches option 3.