`y = 2arccos(x) - 2sqrt(1-x^2)` Find the derivative of the function

Friday, November 18, 2011

$\displaystyle{y}={2}{\arccos{{\left({x}\right)}}}-{2}\sqrt{{{1}-{{x}}^{{2}}}}$ Find the derivative of the function

$\displaystyle{y}={2}{\arccos{{\left({x}\right)}}}-{2}\sqrt{{{1}-{{x}}^{{2}}}}$

First, express the radical in exponent form.

$\displaystyle{y}={2}{\arccos{{\left({x}\right)}}}-{2}{{\left({1}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}$

To take the derivative of this, use the following formulas:

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({\arccos{{\left({u}\right)}}}\right)}=-\frac{{1}}{\sqrt{{{1}-{{u}}^{{2}}}}}\frac{{{d}{u}}}{{\left.{d}{x}}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({{u}}^{{n}}\right)}={n}\cdot{{u}}^{{{n}-{1}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

Applying these formulas, the derivative of the function will be:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{d}}{{\left.{d}{x}}}{\left[{2}{\arccos{{\left({x}\right)}}}-{2}{{\left({1}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{d}}{{\left.{d}{x}}}{\left[{2}{\arccos{{\left({x}\right)}}}\right]}-\frac{{d}}{{\left.{d}{x}}}{\left[{2}{{\left({1}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}={2}\frac{{d}}{{\left.{d}{x}}}{\left[{\arccos{{\left({x}\right)}}}\right]}-{2}\frac{{d}}{{\left.{d}{x}}}{\left[{{\left({1}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}={2}\cdot{\left(-\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}\right)}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({x}\right)}-{2}\cdot\frac{{1}}{{2}}{{\left({1}-{{x}}^{{2}}\right)}}^{{-\frac{{1}}{{2}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({1}-{{x}}^{{2}}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}={2}\cdot{\left(-\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}\right)}\cdot{1}-{2}\cdot\frac{{1}}{{2}}{{\left({1}-{{x}}^{{2}}\right)}}^{{-\frac{{1}}{{2}}}}\cdot{\left(-{2}{x}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}={2}\cdot{\left(-\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}\right)}\cdot{1}-{2}\cdot\frac{{1}}{{2}}\cdot\frac{{1}}{{{{\left({1}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}}}\cdot{\left(-{2}{x}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}={2}\cdot{\left(-\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}\right)}\cdot{1}-{2}\cdot\frac{{1}}{{2}}\cdot\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}\cdot{\left(-{2}{x}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=-\frac{{2}}{\sqrt{{{1}-{{x}}^{{2}}}}}+\frac{{{2}{x}}}{\sqrt{{{1}-{{x}}^{{2}}}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{{2}{x}-{2}}}{\sqrt{{{1}-{{x}}^{{2}}}}}$

Therefore, the derivative of the function is $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{{2}{x}-{2}}}{\sqrt{{{1}-{{x}}^{{2}}}}}$ .

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Friday, November 18, 2011

$\displaystyle{y}={2}{\arccos{{\left({x}\right)}}}-{2}\sqrt{{{1}-{{x}}^{{2}}}}$ Find the derivative of the function

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Friday, November 18, 2011

Find the derivative of the function

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Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{y}={2}{\arccos{{\left({x}\right)}}}-{2}\sqrt{{{1}-{{x}}^{{2}}}}$ Find the derivative of the function

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?