Q. 32 - 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.   

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

Step-by-Step Solution

Answer

The required answer is $-\frac{3}{4\sqrt{3x+1}\sqrt{2-\sqrt{3x+1}}}$

Step 1. Given Information

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

Step 2. Calculation

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

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