`y = arctan(x/2) - 1/(2(x^2+4))` Find the derivative of the function

Tuesday, April 17, 2012

$\displaystyle{y}={\arctan{{\left(\frac{{x}}{{2}}\right)}}}-\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}$ Find the derivative of the function

The derivative of y with respect to is denoted as $\displaystyle{y}'$ or $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ .

For the given equation: $\displaystyle{y}={\arctan{{\left(\frac{{x}}{{2}}\right)}}}-\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}$ ,

we may apply the basic property of derivative:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({u}-{v}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({u}\right)}-\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({v}\right)}$

Then the derivative of y will be:

$\displaystyle{y}'=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\arctan{{\left(\frac{{x}}{{2}}\right)}}}-\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}\right)}$

$\displaystyle{y}'=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\arctan{{\left(\frac{{x}}{{2}}\right)}}}\right)}-\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}\right)}$

To find the derivative of the first term: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\arctan{{\left(\frac{{x}}{{2}}\right)}}}\right)}$ , recall the basic derivative formula for inverse tangent as:

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

With $\displaystyle{u}=\frac{{x}}{{2}}$ and $\displaystyle{d}{u}={\left(\frac{{1}}{{2}}\right)}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}=\frac{{1}}{{2}}$ , we will have:

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

$\displaystyle=\frac{{\frac{{1}}{{2}}}}{{{1}+{\left(\frac{{{x}}^{{2}}}{{4}}\right)}}}$

Express the bottom as one fraction:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\arctan{{\left(\frac{{x}}{{2}}\right)}}}\right)}=\frac{{\frac{{1}}{{2}}}}{{\frac{{{{x}}^{{2}}+{4}}}{{4}}}}$

Flip the bottom to proceed to multiplication:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\arctan{{\left(\frac{{x}}{{2}}\right)}}}\right)}=\frac{{1}}{{2}}\cdot\frac{{4}}{{{{x}}^{{2}}+{4}}}$

$\displaystyle=\frac{{4}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}$

$\displaystyle=\frac{{2}}{{{{x}}^{{2}}+{4}}}$

For the derivative of the second term: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}\right)}$ , we can rewrite it using the basic property of derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({c}\cdot{f{{\left({x}\right)}}}\right)}={c}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{f{{\left({x}\right)}}}$ where c is constant.

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}\right)}={\left(\frac{{1}}{{2}}\right)}\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{{x}}^{{2}}+{4}}}\right)}$

Then apply the Quotient Rule for derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{u}}{{v}}\right)}=\frac{{{u}'\cdot{v}-{v}'\cdot{u}}}{{{v}}^{{2}}}$ on $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}\right)}$ .

We let:

$\displaystyle{u}={1}$ then $\displaystyle{u}'={0}$

$\displaystyle{v}={{x}}^{{2}}+{4}$ then $\displaystyle{v}'={2}{x}$

$\displaystyle{{v}}^{{2}}={{\left({{x}}^{{2}}+{4}\right)}}^{{2}}$

Applying the Quotient rule, we get:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}\right)},=\frac{{{0}\cdot{\left({{x}}^{{2}}+{4}\right)}-{\left({1}\right)}{\left({2}{x}\right)}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$

$\displaystyle=\frac{{{0}-{2}{x}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$

$\displaystyle=\frac{{-{2}{x}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$

Then $\displaystyle{\left(\frac{{1}}{{2}}\right)}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{{x}}^{{2}}+{4}}}\right)}={\left(\frac{{1}}{{2}}\right)}\cdot\frac{{-{2}{x}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$

$\displaystyle=-\frac{{x}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$

For the complete problem:

$\displaystyle{y}'=\frac{{2}}{{{{x}}^{{2}}+{4}}}+\frac{{x}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$

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Tuesday, April 17, 2012

$\displaystyle{y}={\arctan{{\left(\frac{{x}}{{2}}\right)}}}-\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}$ 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?

Tuesday, April 17, 2012

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?

$\displaystyle{y}={\arctan{{\left(\frac{{x}}{{2}}\right)}}}-\frac{{1}}{{{2}{\left({{x}}^{{2}}+{4}\right)}}}$ 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?