`f(x) = arcsec(2x)` Find the derivative of the function

Saturday, November 22, 2014

$\displaystyle{f{{\left({x}\right)}}}={a}{r}{c}{\sec{{\left({2}{x}\right)}}}$ Find the derivative of the function

The given function: $\displaystyle{f{{\left({x}\right)}}}={a}{r}{c}{\sec{{\left({2}{x}\right)}}}$ is in a form of an inverse trigonometric function.

For the derivative formula of an inverse secant function, we follow:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({a}{r}{c}{\sec{{\left({u}\right)}}}\right)}=\frac{{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}}{{{\left|{u}\right|}\sqrt{{{{u}}^{{2}}-{1}}}}}$

To be able to apply the formula, we let u $\displaystyle={2}{x}$ then $\displaystyle{{u}}^{{2}}={{\left({2}{x}\right)}}^{{2}}={4}{{x}}^{{2}}$ and

$\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}={2}$ .

It follows that $\displaystyle{f{{\left({x}\right)}}}={a}{r}{c}{\sec{{\left({2}{x}\right)}}}$ will have a derivative of:

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

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{2}}{{{\left|{2}{x}\right|}\sqrt{{{4}{{x}}^{{2}}-{1}}}}}$

Cancel out common factor 2 from top and bottom:

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{{\left|{x}\right|}\sqrt{{{4}{{x}}^{{2}}-{1}}}}}$

This can also be written as : $\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{\sqrt{{{{x}}^{{2}}}}\sqrt{{{4}{{x}}^{{2}}-{1}}}}}$ since

$\displaystyle{\left|{x}\right|}=\sqrt{{{{x}}^{{2}}}}$

Then applying the radical property: $\displaystyle\sqrt{{{a}}}\cdot\sqrt{{{b}}}=\sqrt{{{a}\cdot{b}}}$ at the bottom, we get:

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{\sqrt{{{{x}}^{{2}}\cdot{\left({4}{{x}}^{{2}}-{1}\right)}}}}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{\sqrt{{{4}{{x}}^{{4}}-{{x}}^{{2}}}}}$

The derivative of the function f(x) =arcsec(x) can be :

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{{\left|{x}\right|}\sqrt{{{4}{{x}}^{{2}}-{1}}}}}$

or $\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{\sqrt{{{{x}}^{{2}}}}\sqrt{{{4}{{x}}^{{2}}-{1}}}}}$

or $\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{\sqrt{{{4}{{x}}^{{4}}-{{x}}^{{2}}}}}}$

Blog

Saturday, November 22, 2014

$\displaystyle{f{{\left({x}\right)}}}={a}{r}{c}{\sec{{\left({2}{x}\right)}}}$ Find the derivative of the function

No comments:

Post a Comment

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

Saturday, November 22, 2014

Find the derivative of the function

No comments:

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?

$\displaystyle{f{{\left({x}\right)}}}={a}{r}{c}{\sec{{\left({2}{x}\right)}}}$ 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?