`f(x) = 2arcsin(x-1)` Find the derivative of the function

Sunday, August 17, 2014

$\displaystyle{f{{\left({x}\right)}}}={2}{\arcsin{{\left({x}-{1}\right)}}}$ Find the derivative of the function

The derivative of function f with respect to x is denoted as $\displaystyle{f{'}}{\left({x}\right)}$ .

To take the derivative of the given function: $\displaystyle{f{{\left({x}\right)}}}={2}{\arcsin{{\left({x}-{1}\right)}}}$ ,

we can apply the basic property: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left[{c}\cdot{f{{\left({x}\right)}}}\right]}={c}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\left[{f{{\left({x}\right)}}}\right]}$ .

then $\displaystyle{f{'}}{\left({x}\right)}={2}\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\arcsin{{\left({x}-{1}\right)}}}\right)}$

To solve for the $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\arcsin{{\left({x}-{1}\right)}}}\right)}$ , we consider the derivative formula of an inverse trigonometric function.

For the derivative of inverse "sine" function, we follow:

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

To apply the formula with the given function, we let $\displaystyle{u}={x}-{1}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}={1}$ .

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

Then $\displaystyle{f{'}}{\left({x}\right)}={2}\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\arcsin{{\left({x}-{1}\right)}}}\right)}$ becomes:

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

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

To further simplify, we can evaluate the exponent inside the radical:

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

Note: $\displaystyle{{\left({x}-{1}\right)}}^{{2}}={\left({x}-{1}\right)}{\left({x}-{1}\right)}$

Applying FOIL or distributive property:

$\displaystyle{\left({x}-{1}\right)}{\left({x}-{1}\right)}={x}\cdot{x}+{x}\cdot{\left(-{1}\right)}+{\left(-{1}\right)}\cdot{x}+{\left(-{1}\right)}{\left(-{1}\right)}$

$\displaystyle={{x}}^{{2}}–{x}–{x}+{1}$

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

Simplify the expression inside radical:

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

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{2}}{\sqrt{{-{{x}}^{{2}}+{2}{x}}}}{\quad\text{or}\quad}{f{'}}{\left({x}\right)}=\frac{{2}}{\sqrt{{{2}{x}-{{x}}^{{2}}}}}$

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Sunday, August 17, 2014

$\displaystyle{f{{\left({x}\right)}}}={2}{\arcsin{{\left({x}-{1}\right)}}}$ 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?

Sunday, August 17, 2014

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?

$\displaystyle{f{{\left({x}\right)}}}={2}{\arcsin{{\left({x}-{1}\right)}}}$ 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?