Derivative of $$$\sqrt{a^{x} - 1}$$$ with respect to $$$x$$$

The calculator will find the derivative of $$$\sqrt{a^{x} - 1}$$$ with respect to $$$x$$$, with steps shown.

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Find $$$\frac{d}{dx} \left(\sqrt{a^{x} - 1}\right)$$$.

Solution

The function $$$\sqrt{a^{x} - 1}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \sqrt{u}$$$ and $$$g{\left(x \right)} = a^{x} - 1$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\sqrt{a^{x} - 1}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right) \frac{d}{dx} \left(a^{x} - 1\right)\right)}$$

Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = \frac{1}{2}$$$:

$${\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right)\right)} \frac{d}{dx} \left(a^{x} - 1\right) = {\color{red}\left(\frac{1}{2 \sqrt{u}}\right)} \frac{d}{dx} \left(a^{x} - 1\right)$$

Return to the old variable:

$$\frac{\frac{d}{dx} \left(a^{x} - 1\right)}{2 \sqrt{{\color{red}\left(u\right)}}} = \frac{\frac{d}{dx} \left(a^{x} - 1\right)}{2 \sqrt{{\color{red}\left(a^{x} - 1\right)}}}$$

The derivative of a sum/difference is the sum/difference of derivatives:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(a^{x} - 1\right)\right)}}{2 \sqrt{a^{x} - 1}} = \frac{{\color{red}\left(\frac{d}{dx} \left(a^{x}\right) - \frac{d}{dx} \left(1\right)\right)}}{2 \sqrt{a^{x} - 1}}$$

Apply the exponential rule $$$\frac{d}{dx} \left(n^{x}\right) = n^{x} \ln\left(n\right)$$$ with $$$n = a$$$:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(a^{x}\right)\right)} - \frac{d}{dx} \left(1\right)}{2 \sqrt{a^{x} - 1}} = \frac{{\color{red}\left(a^{x} \ln\left(a\right)\right)} - \frac{d}{dx} \left(1\right)}{2 \sqrt{a^{x} - 1}}$$

The derivative of a constant is $$$0$$$:

$$\frac{a^{x} \ln\left(a\right) - {\color{red}\left(\frac{d}{dx} \left(1\right)\right)}}{2 \sqrt{a^{x} - 1}} = \frac{a^{x} \ln\left(a\right) - {\color{red}\left(0\right)}}{2 \sqrt{a^{x} - 1}}$$

Thus, $$$\frac{d}{dx} \left(\sqrt{a^{x} - 1}\right) = \frac{a^{x} \ln\left(a\right)}{2 \sqrt{a^{x} - 1}}$$$.

Answer

$$$\frac{d}{dx} \left(\sqrt{a^{x} - 1}\right) = \frac{a^{x} \ln\left(a\right)}{2 \sqrt{a^{x} - 1}}$$$A


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