Blog: `y = ln(sqrt(x^2 - 4))` Find the derivative of the function.

$\displaystyle{y}={\ln{{\left(\sqrt{{{{x}}^{{2}}-{4}}}\right)}}}$

First, use the formula:

$\displaystyle{\left({\ln{{u}}}\right)}'=\frac{{1}}{{u}}\cdot{u}'$

Applying that formula, the derivative of the function will be:

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

To take the derivative of the inner function, express the radical in exponent form.

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

Then, use the formula:

$\displaystyle{\left({{u}}^{{n}}\right)}'={n}\cdot{{u}}^{{{n}-{1}}}\cdot{u}'$

So, y' will become:

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

To take the derivative of the innermost function, use the formulas:

$\displaystyle{\left({{x}}^{{n}}\right)}'={n}\cdot{{x}}^{{{n}-{1}}}$

$\displaystyle{\left({c}\right)}'={0}$

Applying these two formulas, y' will become:

$\displaystyle{y}'=\frac{{1}}{\sqrt{{{{x}}^{{2}}-{4}}}}\cdot\frac{{1}}{{2}}{{\left({{x}}^{{2}}-{4}\right)}}^{{-\frac{{1}}{{2}}}}\cdot{\left({2}{x}-{0}\right)}$

Simplifying...

$\displaystyle{y}={\ln{{\left(\sqrt{{{{x}}^{{2}}-{4}}}\right)}}}$

First, use the formula:

$\displaystyle{\left({\ln{{u}}}\right)}'=\frac{{1}}{{u}}\cdot{u}'$

Applying that formula, the derivative of the function will be:

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

To take the derivative of the inner function, express the radical in exponent form.

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

Then, use the formula:

$\displaystyle{\left({{u}}^{{n}}\right)}'={n}\cdot{{u}}^{{{n}-{1}}}\cdot{u}'$

So, y' will become:

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

To take the derivative of the innermost function, use the formulas:

$\displaystyle{\left({{x}}^{{n}}\right)}'={n}\cdot{{x}}^{{{n}-{1}}}$

$\displaystyle{\left({c}\right)}'={0}$

Applying these two formulas, y' will become:

$\displaystyle{y}'=\frac{{1}}{\sqrt{{{{x}}^{{2}}-{4}}}}\cdot\frac{{1}}{{2}}{{\left({{x}}^{{2}}-{4}\right)}}^{{-\frac{{1}}{{2}}}}\cdot{\left({2}{x}-{0}\right)}$

Simplifying it will result to:

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

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

$\displaystyle{y}'=\frac{{1}}{\sqrt{{{{x}}^{{2}}-{4}}}}\cdot\frac{{1}}{{2}}\cdot\frac{{1}}{\sqrt{{{{x}}^{{2}}-{4}}}}\cdot{2}{x}$

$\displaystyle{y}'=\frac{{x}}{{{{x}}^{{2}}-{4}}}$

Therefore, the derivative of the given function is $\displaystyle{y}'=\frac{{x}}{{{{x}}^{{2}}-{4}}}$ .

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$\displaystyle{y}={\ln{{\left(\sqrt{{{{x}}^{{2}}-{4}}}\right)}}}$ Find the derivative of the function.

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