Exponential Differentiation

We are going to focus on differentiating functions of the form \( f(x)^{g(x)} \), where both the base and the exponent are functions of \( x \).

When the power is constant, we already know how to differentiate such cases. For example,

\[ \frac{d}{dx} (1 + x^2)^{3/2} = \frac{3}{2} (1 + x^2)^{1/2} \cdot 2x = 3x (1 + x^2)^{1/2}. \]

When the base is constant but the exponent is variable, we differentiate using the standard exponential differentiation rule. For example,

\[ \frac{d}{dx} 4^{\sin x} = 4^{\sin x} \ln 4 \cos x. \]

However, when both the base and the power are variable, we cannot directly apply the above rules. The differentiation of such functions requires a different approach.

To differentiate \( f(x)^{g(x)} \), we begin by rewriting it using the fundamental identity

\[ a^b = e^{b \ln a}. \]

Applying this,

\[ f(x)^{g(x)} = e^{g(x) \ln f(x)}. \]

This transformation allows us to differentiate using the chain rule, rather than dealing with the exponent directly. Taking derivatives on both sides,

\[ \frac{d}{dx} f(x)^{g(x)} = \frac{d}{dx} e^{g(x) \ln f(x)}. \]

Since the derivative of \( e^u \) is \( e^u \cdot \frac{du}{dx} \), applying this to our expression gives

\[ = e^{g(x) \ln f(x)} \cdot \frac{d}{dx} (g(x) \ln f(x)). \]

Since we earlier rewrote \( f(x)^{g(x)} \) as \( e^{g(x) \ln f(x)} \), we substitute it back:

\[ = f(x)^{g(x)} \cdot \frac{d}{dx} (g(x) \ln f(x)). \]

To differentiate \( g(x) \ln f(x) \), we use the product rule. The first term, \( g(x) \), is multiplied by \( \ln f(x) \), so differentiating each while treating the other as a constant,

\[ \frac{d}{dx} (g(x) \ln f(x)) = g(x) \frac{d}{dx} \ln f(x) + \ln f(x) \frac{d}{dx} g(x). \]

Using the derivative \( \frac{d}{dx} \ln f(x) = \frac{f'(x)}{f(x)} \), we get

\[ = g(x) \frac{f'(x)}{f(x)} + \ln f(x) g'(x). \]

Substituting this back into our main equation,

\[ \frac{d}{dx} f(x)^{g(x)} = f(x)^{g(x)} \left( g(x) \frac{f'(x)}{f(x)} + \ln f(x) g'(x) \right). \]

This expression gives the derivative of \( f(x)^{g(x)} \), capturing both the dependence of \( f(x) \) and \( g(x) \) on \( x \).

We need not remember this as a formula. Try to understand the process only.

Let us apply this method to the function \( f(x) = x^x \). Rewriting it,

\[ x^x = e^{x \ln x}. \]

Now differentiating both sides,

\[ \frac{d}{dx} x^x = \frac{d}{dx} e^{x \ln x} = e^{x \ln x} \cdot \frac{d}{dx} (x \ln x). \]

Since \( e^{x \ln x} = x^x \), substituting back,

\[ = x^x \cdot \frac{d}{dx} (x \ln x) = x^x \cdot (x \frac{d}{dx} \ln x + \ln x \frac{d}{dx} x). \]

\[ = x^x \cdot (1 + \ln x). \]

Thus,

\[ \frac{d}{dx} x^x = x^x (1 + \ln x). \]

This follows directly from our logarithmic transformation, making differentiation of variable exponents systematic.

Example

Differentiate \( f(x) = (\sin x)^x \).

Solution:

Rewriting the function using the identity \( a^b = e^{b \ln a} \),

\[ (\sin x)^x = e^{x \ln (\sin x)}. \]

Now differentiating both sides,

\[ \frac{d}{dx} (\sin x)^x = \frac{d}{dx} e^{x \ln (\sin x)} = e^{x \ln (\sin x)} \cdot \frac{d}{dx} (x \ln (\sin x)). \]

Since \( e^{x \ln (\sin x)} = (\sin x)^x \), substituting back,

\[ = (\sin x)^x \cdot \frac{d}{dx} (x \ln (\sin x)). \]

Applying the product rule,

\[ = (\sin x)^x \cdot \left( x \frac{d}{dx} \ln (\sin x) + \ln (\sin x) \frac{d}{dx} x \right). \]

\[ = (\sin x)^x \cdot \left( x \cdot \frac{\cos x}{\sin x} + \ln (\sin x) \right). \]

\[ = (\sin x)^x \cdot \left( x \cot x + \ln (\sin x) \right). \]

Thus,

\[ \frac{d}{dx} (\sin x)^x = (\sin x)^x \left( x \cot x + \ln (\sin x) \right). \]

Example

Differentiate \( x^y + y^x = xy \).

Solution:

Rewriting using exponentials,

\[ e^{y \ln x} + e^{x \ln y} = xy. \]

Differentiating both sides,

\[ e^{y \ln x} \left( \ln x \frac{dy}{dx} + \frac{y}{x} \right) + e^{x \ln y} \left( \ln y + \frac{x}{y} \frac{dy}{dx} \right) = x \frac{dy}{dx} + y. \]

Expanding,

\[ x^y \left( \ln x \frac{dy}{dx} + \frac{y}{x} \right) + y^x \left( \ln y + \frac{x}{y} \frac{dy}{dx} \right) = x \frac{dy}{dx} + y. \]

Rearranging,

\[ x^y \ln x \frac{dy}{dx} + y^x \frac{x}{y} \frac{dy}{dx} - x \frac{dy}{dx} = - x^y \frac{y}{x} - y^x \ln y + y. \]

Factoring \( \frac{dy}{dx} \),

\[ \left( x^y \ln x + \frac{x}{y} y^x - x \right) \frac{dy}{dx} = - x^y \frac{y}{x} - y^x \ln y + y. \]

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