`h(x) = log_3(xsqrt(x-1)/2)` Find the derivative of the function

Tuesday, September 13, 2011

$\displaystyle{h}{\left({x}\right)}={\log}_{{3}}{\left({x}\frac{\sqrt{{{x}-{1}}}}{{2}}\right)}$ Find the derivative of the function

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

To solve for $\displaystyle{h}'{\left({x}\right)}$ for given function $\displaystyle{h}{\left({x}\right)}={\log}_{{3}}{\left({x}\cdot\frac{\sqrt{{{x}-{1}}}}{{2}}\right)}$ , we apply the derivative for logarithmic functions: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\log}_{{a}}{\left({u}\right)}=\frac{{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}}{{{u}\cdot{\ln{{\left({a}\right)}}}}}$ .

We may let $\displaystyle{u}={x}\cdot\frac{\sqrt{{{x}-{1}}}}{{2}}$ and $\displaystyle{a}={3}$ .

In solving for the derivative of u: $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$ , we apply basic property: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({c}\cdot{f{{\left({x}\right)}}}\right)}={c}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{f{{\left({x}\right)}}}$ .

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

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

Applying the Product Rule for derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({f{\cdot}}{g}\right)}={f{'}}\cdot{g{+}}{f{\cdot}}{g{'}}$ .

Let:

$\displaystyle{f{=}}{x}$ then $\displaystyle{f{'}}={1}$

$\displaystyle{g{=}}\sqrt{{{x}-{1}}}$ then $\displaystyle{g{'}}=\frac{{1}}{{{2}\sqrt{{{x}-{1}}}}}$

Then following the formula: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({f{\cdot}}{g}\right)}={f{'}}\cdot{g{+}}{f{\cdot}}{g{'}}$ , we set-up:

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

$\displaystyle{u}'={\left(\frac{{1}}{{2}}\right)}\cdot{\left[{1}\cdot{\left(\sqrt{{{x}-{1}}}+{\left({x}\right)}\cdot{\left(\frac{{1}}{{{2}\sqrt{{{x}-{1}}}}}\right)}\right]}\right.}$

Simplify:

$\displaystyle{u}'={\left(\frac{{1}}{{2}}\right)}\cdot{\left[{1}\cdot{\left(\sqrt{{{x}-{1}}}\cdot\frac{{{2}\sqrt{{{x}-{1}}}}}{{{2}\sqrt{{{x}-{1}}}}}+\frac{{x}}{{{2}\sqrt{{{x}-{1}}}}}\right]}\right.}$

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

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

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

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

$\displaystyle={\left(\frac{{1}}{{2}}\right)}\cdot{\left[\frac{{{3}{x}-{2}}}{{{2}\sqrt{{{x}-{1}}}}}\right]}$

$\displaystyle=\frac{{{3}{x}-{2}}}{{{4}\sqrt{{{x}-{1}}}}}$

Applying $\displaystyle{u}={x}\cdot\frac{\sqrt{{{x}-{1}}}}{{2}}$ , $\displaystyle{a}={3}$ , and $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}{\quad\text{or}\quad}{u}'=\frac{{{3}{x}-{2}}}{{{4}\sqrt{{{x}-{1}}}}}$

with the derivative formula: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\log}_{{a}}{\left({u}\right)}=\frac{{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}}{{{u}\cdot{\ln{{\left({a}\right)}}}}}$ , we get:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\log}_{{3}}{\left({x}\cdot\frac{\sqrt{{{x}-{1}}}}{{2}}\right)}\right)}=\frac{{\frac{{{3}{x}-{2}}}{{{4}\sqrt{{{x}-{1}}}}}}}{{{x}\frac{\sqrt{{{x}-{1}}}}{{2}}{\ln{{\left({3}\right)}}}}}$

This is also the same as:

$\displaystyle{h}'{\left({x}\right)}=\frac{{{3}{x}-{2}}}{{{4}\sqrt{{{x}-{1}}}}}\cdot\frac{{1}}{{\frac{{{x}\sqrt{{{x}-{1}}}{\ln{{\left({3}\right)}}}}}{{2}}}}$

Flip the $\displaystyle{x}\sqrt{{{x}-{1}}}{\ln{{\left({3}\right)}}}\frac{}}{{2}}$ to proceed to multiplication:

$\displaystyle{h}'{\left({x}\right)}=\frac{{{3}{x}-{2}}}{{{4}\sqrt{{{x}-{1}}}}}\cdot{\left({1}\cdot\frac{{2}}{{{x}\frac{\sqrt{{{x}-{1}}}}{{\ln{{\left({3}\right)}}}}}}\right)}$

Multiply across:

$\displaystyle{h}'{\left({x}\right)}=\frac{{{2}{\left({3}{x}-{2}\right)}}}{{{4}{x}{{\left(\sqrt{{{x}-{1}}}\right)}}^{{2}}{\ln{{\left({3}\right)}}}}}$

Simplify by applying: $\displaystyle\frac{{2}}{{4}}=\frac{{1}}{{2}}$ and $\displaystyle{{\left(\sqrt{{{x}-{1}}}\right)}}^{{2}}={x}-{1}$ .

FINAL ANSWER:

$\displaystyle{h}'{\left({x}\right)}=\frac{{{3}{x}-{2}}}{{{2}{x}{\left({x}-{1}\right)}{\ln{{\left({3}\right)}}}}}$

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Tuesday, September 13, 2011

$\displaystyle{h}{\left({x}\right)}={\log}_{{3}}{\left({x}\frac{\sqrt{{{x}-{1}}}}{{2}}\right)}$ 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?

Tuesday, September 13, 2011

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?

$\displaystyle{h}{\left({x}\right)}={\log}_{{3}}{\left({x}\frac{\sqrt{{{x}-{1}}}}{{2}}\right)}$ 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?