Q 39. - Calculus

Question

Find the derivatives of each of functions in Exercises 17–44. In some cases it may be convenient to do some preliminary algebra.

$f\left(x\right)=\sqrt{{\log }_2\left(3^x-5\right)}$

Step-by-Step Solution

Answer

The required answer is $\frac{1}{2}{\left({\log }_2\left(3^x-5\right)\right)}^{-\frac{1}{2}}\left(\frac{1}{\ln 2\left(3^x-5\right)}\left(\ln 3\right)3^x\right)$.

Step 1. Given information.

The given function is:

$f\left(x\right)=\sqrt{{\log }_2\left(3^x-5\right)}$

Step 2. Find the derivative of the given function.

$$\begin{gathered} f\left(x\right)=\sqrt{{\log }_2\left(3^x-5\right)} \\ f\text{'}\left(x\right)=\frac{d}{dx}{\left({\log }_2\left(3^x-5\right)\right)}^{\frac{1}{2}} \\ =\frac{1}{2}{\left({\log }_2\left(3^x-5\right)\right)}^{-\frac{1}{2}}\frac{d}{dx}\left({\log }_2\left(3^x-5\right)\right) \\ =\frac{1}{2}{\left({\log }_2\left(3^x-5\right)\right)}^{-\frac{1}{2}}·\frac{1}{\ln 2\left(3^x-5\right)}·\frac{d}{dx}\left(3^x-5\right) \\ =\frac{1}{2}{\left({\log }_2\left(3^x-5\right)\right)}^{-\frac{1}{2}}·\frac{1}{\ln 2\left(3^x-5\right)}·\ln 3·3^x \\ =\frac{1}{2}{\left({\log }_2\left(3^x-5\right)\right)}^{-\frac{1}{2}}\left(\frac{1}{\ln 2\left(3^x-5\right)}\left(\ln 3\right)3^x\right) \end{gathered}$$