Q. 46 - 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.  

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

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

Step 2. Calculation

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

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