How do you differentiate f(x)=x-sqrt(2^x-x^2) using the chain rule?

Dec 2, 2015

$f ' \left(x\right) = 1 - \frac{{2}^{x} \ln 2 - 2 x}{2 \sqrt{{2}^{x} - {x}^{2}}}$

Explanation:

$f ' \left(x\right) = 1 - \frac{d}{\mathrm{dx}} \left[{\left({2}^{x} - {x}^{2}\right)}^{\frac{1}{2}}\right]$

Find just this derivative through the Chain Rule:

$\frac{d}{\mathrm{dx}} \left[{\left({2}^{x} - {x}^{2}\right)}^{\frac{1}{2}}\right] = \frac{1}{2} {\left({2}^{x} - {x}^{2}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left[{2}^{x} - {x}^{2}\right]$

$= \frac{\frac{d}{\mathrm{dx}} \left[{2}^{x}\right] - 2 x}{2 \sqrt{{2}^{x} - {x}^{2}}}$

Find the remaining derivative.

$\frac{d}{\mathrm{dx}} \left[{2}^{x}\right] = \frac{d}{\mathrm{dx}} \left[{e}^{\ln {2}^{x}}\right] = \frac{d}{\mathrm{dx}} \left[{e}^{x \ln 2}\right] = {e}^{x \ln 2} \frac{d}{\mathrm{dx}} \left[x \ln 2\right] = {2}^{x} \ln 2$

Plug back in:

$\frac{d}{\mathrm{dx}} \left[{\left({2}^{x} - {x}^{2}\right)}^{\frac{1}{2}}\right] = \frac{{2}^{x} \ln 2 - 2 x}{2 \sqrt{{2}^{x} - {x}^{2}}}$

Plug in again.

$f ' \left(x\right) = 1 - \frac{{2}^{x} \ln 2 - 2 x}{2 \sqrt{{2}^{x} - {x}^{2}}}$

(Optional simplification:)

$f ' \left(x\right) = \frac{2 \sqrt{{2}^{x} - {x}^{2}} - {2}^{x} \ln 2 + 2 x}{2 \sqrt{{2}^{x} - {x}^{2}}}$