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

Thursday, May 20, 2010

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

The derivative of y in terms of x is denoted by $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ or $\displaystyle{y}’'$

For the given problem: $\displaystyle{y}={x}{\arctan{{\left({2}{x}\right)}}}-\frac{{1}}{{4}}{\ln{{\left({1}+{4}{{x}}^{{2}}\right)}}}$ , we may apply the basic differentiation property:

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

Then the derivative of the function can be set-up as:

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

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

For the derivative of $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left[{x}{\arctan{{\left({2}{x}\right)}}}\right.}$ , we apply the Product Rule: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({u}\cdot{v}\right)}={u}’\cdot{v}=+{u}\cdot{v}’$ .

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left[{x}{\arctan{{\left({2}{x}\right)}}}\right]}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({x}\right)}\cdot{\arctan{{\left({2}{x}\right)}}}+{x}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\arctan{{\left({2}{x}\right)}}}$ .

Let $\displaystyle{u}={x}$ then $\displaystyle{u}'={1}$

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

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

Then,

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

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

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

For the derivative of $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}\frac{{1}}{{4}}{\ln{{\left({1}+{4}{{x}}^{{2}}\right)}}}$ , we apply the basic derivative property:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{c}\cdot{f{{\left({x}\right)}}}={c}\frac{{d}}{{{\left.{d}{x}\right.}}}{f{{\left({x}\right)}}}$ .

Then,

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

Apply the basic derivative formula for natural logarithm function: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\ln{{\left({u}\right)}}}=\frac{{{d}{u}}}{{u}}$ .

Let $\displaystyle{u}={1}+{4}{{x}}^{{2}}$ then $\displaystyle{d}{u}={8}{x}$

$\displaystyle\frac{{1}}{{4}}\frac{{d}}{{{\left.{d}{x}\right.}}}{\ln{{\left({1}+{4}{{x}}^{{2}}\right)}}}=\frac{{1}}{{4}}\cdot{8}\frac{{x}}{{{1}+{4}{{x}}^{{2}}}}$

$\displaystyle=\frac{{{2}{x}}}{{{1}+{4}{{x}}^{{2}}}}$

Combining the results, we get:

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

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

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

$\displaystyle{y}'={\arctan{{\left({2}{x}\right)}}}+{0}$

$\displaystyle{y}'={\arctan{{\left({2}{x}\right)}}}$

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Thursday, May 20, 2010

$\displaystyle{y}={x}{\arctan{{\left({2}{x}\right)}}}-\frac{{1}}{{4}}{\ln{{\left({1}+{4}{{x}}^{{2}}\right)}}}$ 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, May 20, 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?

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