`y = tanh^-1(sqrt(x))` Find the derivative of the function

Tuesday, May 5, 2009

$\displaystyle{y}={{\tanh}}^{{-{{1}}}}{\left(\sqrt{{{x}}}\right)}$ Find the derivative of the function

Derivative of a function f with respect to x is denoted as $\displaystyle{f{'}}{\left({x}\right)}$ or $\displaystyle{y}'$ .

To solve for derivative of y or $\displaystyle{\left({y}'\right)}$ for the given problem: $\displaystyle{y}={{\tanh}}^{{-{1}}}{\left(\sqrt{{{x}}}\right)}$ , we follow the basic derivative formula for inverse hyperbolic function:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({{\tanh}}^{{-{1}}}{\left({u}\right)}\right)}=\frac{{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}}{{{1}-{{u}}^{{2}}}}$ where $\displaystyle{\left|{u}\right|}\lt{1}$ .

Let: $\displaystyle{u}=\sqrt{{{x}}}$

Apply the Law of Exponent: $\displaystyle\sqrt{{{x}}}={{x}}^{{\frac{{1}}{{2}}}}$

Solve for the derivative of u using the Power Rule for derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{n}}={n}\cdot{{x}}^{{{n}+{1}}}\cdot{d}{\left({x}\right)}$

Then,

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

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

Apply the Law of Exponent:

$\displaystyle{{x}}^{{-{n}}}=\frac{{1}}{{{x}}^{{n}}}.$

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

Rearrange into:

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

$\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}=\frac{{1}}{{{2}\sqrt{{{x}}}}}$

Apply the derivative formula, we get:

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

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

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

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

Final answer:

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

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Tuesday, May 5, 2009

$\displaystyle{y}={{\tanh}}^{{-{{1}}}}{\left(\sqrt{{{x}}}\right)}$ 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?

Tuesday, May 5, 2009

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?

$\displaystyle{y}={{\tanh}}^{{-{{1}}}}{\left(\sqrt{{{x}}}\right)}$ 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?