Q. 35 - Calculus

Question

Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating.  

$f\left(x\right)=\sqrt{3x-4{(2x+1)}^6}$

Step-by-Step Solution

Answer

The required answer is $\frac{3-24{(2x+1)}^5\left(2\right)}{2\sqrt{3x-4{(2x+1)}^6}}$

Step 1. Given Information

The given function is   $f\left(x\right)=\sqrt{3x-4{(2x+1)}^6}$

Step 2. Calculation

Differentiate both the sides with respect to x, we get,    

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}{\left(3x-4{(2x+1)}^6\right)}^{\frac{1}{2}} \\ =\frac{1}{2}{\left(3x-4{(2x+1)}^6\right)}^{\frac{1}{2}-1}\frac{d}{dx}\left(\left(3x-4{(2x+1)}^6\right)\right) \\ =\frac{1}{2}{\left(3x-4{(2x+1)}^6\right)}^{\frac{-1}{2}}\left(3-4\left(6{\left(2x+1\right)}^{6-1}\frac{d}{dx}\left(2x+1\right)\right)\right) \\ =\frac{1}{2}{\left(3x-4{(2x+1)}^6\right)}^{\frac{-1}{2}}\left(3-24{\left(2x+1\right)}^{6-1}\left(2\right)\right) \\ =\frac{3-24{\left(2x+1\right)}^5}{\sqrt{3x-4{(2x+1)}^6}} \end{gathered}$$