`f(x) = arctan(sqrt(x))` Find the derivative of the function

Thursday, November 4, 2010

$\displaystyle{f{{\left({x}\right)}}}={\arctan{{\left(\sqrt{{{x}}}\right)}}}$ Find the derivative of the function

The given function $\displaystyle{f{{\left({x}\right)}}}={\arctan{{\left(\sqrt{{{x}}}\right)}}}$ is in a inverse trigonometric form.

The basic derivative formula for inverse tangent is:

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

Using u-substitution, let $\displaystyle{u}=\sqrt{{{x}}}$ then
$\displaystyle{{u}}^{{2}}={{\left(\sqrt{{{x}}}\right)}}^{{2}}={x}$

For the derivative of u or $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$ , we apply the Power Rule:

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

This is possible since $\displaystyle\sqrt{{{x}}}={{x}}^{{\frac{{1}}{{2}}}}$

Then $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{\frac{{1}}{{2}}}}=\frac{{1}}{{2}}\cdot{{x}}^{{\frac{{1}}{{2}}-{1}}}\cdot{1}$ Note: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({x}\right)}={1}$

$\displaystyle=\frac{{1}}{{2}}{{x}}^{{-\frac{{1}}{{2}}}}$

Applying the Law of Exponents: $\displaystyle{{x}}^{{-{n}}}=\frac{{1}}{{{x}}^{{n}}}$ .

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

$\displaystyle=\frac{{1}}{{2}}\cdot\frac{{1}}{{{x}}^{{\frac{{1}}{{2}}}}}$

$\displaystyle=\frac{{1}}{{{2}{{x}}^{{\frac{{1}}{{2}}}}}}{\quad\text{or}\quad}\frac{{1}}{{{2}\sqrt{{{x}}}}}$

We now have: $\displaystyle{u}=\sqrt{{{x}}}$ and $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}=\frac{{1}}{{{2}\sqrt{{{x}}}}}$ .

Applying the formula $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\arctan{{\left({u}\right)}}}=\frac{{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}}{{{1}+{{u}}^{{2}}}}$ :

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\arctan{{\left(\sqrt{{{x}}}\right)}}}=\frac{{\frac{{1}}{{{2}\sqrt{{{x}}}}}}}{{{1}+{{\left(\sqrt{{{x}}}\right)}}^{{2}}}}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{\frac{{1}}{{{2}\sqrt{{{x}}}}}}}{{{1}+{x}}}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{{2}\sqrt{{{x}}}}}\cdot\frac{{1}}{{{1}+{x}}}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{{2}\sqrt{{{x}}}{\left({1}+{x}\right)}}}$

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Thursday, November 4, 2010

$\displaystyle{f{{\left({x}\right)}}}={\arctan{{\left(\sqrt{{{x}}}\right)}}}$ 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?

Thursday, November 4, 2010

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?

$\displaystyle{f{{\left({x}\right)}}}={\arctan{{\left(\sqrt{{{x}}}\right)}}}$ 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?