`h(x) = ln(2x^2 + 1)` Find the derivative of the function.

Thursday, April 5, 2012

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

First, apply the formula:

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

So, the derivative of the function will be:

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

To take the derivative of the inner function, apply the formula:

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

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

So h'(x) will become:

$\displaystyle{h}'{\left({x}\right)}=\frac{{1}}{{{2}{{x}}^{{2}}+{1}}}\cdot{\left({2}\cdot{2}{x}+{0}\right)}$

$\displaystyle{h}'{\left({x}\right)}=\frac{{1}}{{{2}{{x}}^{{2}}+{1}}}\cdot{4}{x}$

$\displaystyle{h}'{\left({x}\right)}=\frac{{{4}{x}}}{{{2}{{x}}^{{2}}+{1}}}$

Therefore, the derivative of the given function is $\displaystyle{h}'{\left({x}\right)}=\frac{{{4}{x}}}{{{2}{{x}}^{{2}}+{1}}}$ .

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