Differentiation of the general exponential function

Let^5.6$y = u^v$. Taking the logarithm of both sides to the base $e$, $\log_e y= v \log_e u$, or, $y = e^{v \log u}$.

Differentiating by formula (IXa),

$$\begin{array}{ll} \frac{dy}{dx} & = e^{v \log u} \frac{d}{d... ...v}{u} \frac{du}{dx} + \log u \frac{dv}{dx} \right ) \end{array}$$

Therefore, $\frac{d}{dx}(u^v) = vu^{v - 1}\frac{du}{dx} + \log u \cdot u^v \frac{dv}{dx}$ (equation (X) in §5.1 above).

The derivative of a function with a variable exponent is equal to the sum of the two results obtained by first differentiating by (VI), regarding the exponent as constant, and again differentiating by (IX), regarding the function as constant.

Let $v = n$, any constant; then (X) reduces to

$$\frac{d}{dx}(u^n) = nu^{n - 1} \frac{du}{dx}.$$

But this is the form differentiated in §5.8; therefore (VI) holds true for any value of $n$.

Example 5.15.1   Differentiate $y = \log(x^2 + a)$.

Solution.

$$\begin{array}{ll} \frac{dy}{dx} &= \frac{\frac{d}{dx}(x^2 + ... ...\ VIIIa}\ (v = x^2 + a)\\ &= \frac{2x}{x^2 + a} . \end{array}$$

Example 5.15.2   Differentiate $y = \log \sqrt{1 - x^2}$.

Solution.

$$\begin{array}{ll} \frac{dy}{dx} & = \frac{\frac{d}{dx}(1 -... ...{1}{2}}} \ \ {\rm by\ VI}\\ & = \frac{x}{x^2 - 1}. \end{array}$$

Example 5.15.3   Differentiate $y = a^{3x^2}$.

Solution.

$$\begin{array}{ll} \frac{dy}{dx} & = \log a \cdot a^{3x^2} \... ... \ \ {\rm by\ IX}\\ & = 6x \log a \cdot a^{3x^2}. \end{array}$$

Example 5.15.4   Differentiate $y = be^{c^2 + x^2}$.

Solution.

$$\begin{array}{ll} \frac{dy}{dx} &= b\frac{d}{dx} \left ( e... ...2 + x^2) \ \ {\rm by\ IXa}\\ & = 2bxe^{c^2 + x^2}. \end{array}$$

Example 5.15.5   Differentiate $y = x^{e^x}$.

Solution.

$$\begin{array}{ll} \frac{dy}{dx} & = e^x x^{e^x - 1} \frac{... ... e^x x^{e^x} \left ( \frac{1}{x} + \log x \right ) \end{array}$$