`y = x^(2/x)` Use logarithmic differentiation to find dy/dx

Sunday, June 13, 2010

$\displaystyle{y}={{x}}^{{\frac{{2}}{{x}}}}$ Use logarithmic differentiation to find dy/dx

For the given problem: $\displaystyle{y}={{x}}^{{\frac{{2}}{{x}}}}$ , we apply the natural logarithm on both sides:

$\displaystyle{\ln{{\left({y}\right)}}}={\ln{{\left({{x}}^{{\frac{{2}}{{x}}}}\right)}}}$

Apply the natural logarithm property: $\displaystyle{\ln{{\left({{x}}^{{n}}\right)}}}={n}\cdot{\ln{{\left({x}\right)}}}$ .

$\displaystyle{\ln{{\left({y}\right)}}}={\left(\frac{{2}}{{x}}\right)}\cdot{\ln{{\left({x}\right)}}}$

Apply chain rule on the left side since y is is function of x.

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({\ln{{\left({y}\right)}}}\right)}=\frac{{1}}{{y}}\cdot{y}'$

Apply product rule: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({u}\cdot{v}\right)}={u}'\cdot{v}+{v}'\cdot{u}$ on the right side:

Let $\displaystyle{u}=\frac{{2}}{{x}}$ then $\displaystyle{u}'=-\frac{{2}}{{{x}}^{{2}}}$

$\displaystyle{v}={\ln{{\left({x}\right)}}}$ then $\displaystyle{v}'=\frac{{1}}{{x}}$

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

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

$\displaystyle=\frac{{-{2}}}{{{{x}}^{{2}}{\ln{{\left({x}\right)}}}}}+\frac{{2}}{{{x}}^{{2}}}$

$\displaystyle=\frac{{-{2}{\ln{{\left({x}\right)}}}+{2}}}{{{x}}^{{2}}}$

The derivative of $\displaystyle{\ln{{\left({y}\right)}}}={\left(\frac{{2}}{{x}}\right)}\cdot{\ln{{\left({x}\right)}}}$ becomes :

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

Isolate y' by multiplying both sides by (y):

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

$\displaystyle{y}'=\frac{{{\left(-{2}{\ln{{\left({x}\right)}}}+{2}\right)}\cdot{y}}}{{{x}}^{{2}}}$

Plug-in $\displaystyle{y}={{x}}^{{\frac{{2}}{{x}}}}$ on the right side:

$\displaystyle{y}'=\frac{{{\left(-{2}{\ln{{\left({x}\right)}}}+{2}\right)}\cdot{{x}}^{{\frac{{2}}{{x}}}}}}{{{x}}^{{2}}}$

Or $\displaystyle{y}'={\left({\left(-{2}{\ln{{\left({x}\right)}}}+{2}\right)}\cdot{{x}}^{{\frac{{2}}{{x}}}}\right)}\cdot{{x}}^{{-{2}}}$

$\displaystyle{y}'={\left(-{2}{\ln{{\left({x}\right)}}}+{2}\right)}\cdot{{x}}^{{\frac{{2}}{{x}}-{2}}}$

$\displaystyle{y}'={\left(-{2}{\ln{{\left({x}\right)}}}+{2}\right)}\cdot{{x}}^{{\frac{{{2}-{2}{x}}}{{x}}}}$

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

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

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Sunday, June 13, 2010

$\displaystyle{y}={{x}}^{{\frac{{2}}{{x}}}}$ Use logarithmic differentiation to find dy/dx

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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, June 13, 2010

Use logarithmic differentiation to find dy/dx

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?

$\displaystyle{y}={{x}}^{{\frac{{2}}{{x}}}}$ Use logarithmic differentiation to find dy/dx

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