Notes: `y = sqrt(x + sqrt(x + sqrt(x)))` Find the derivative of the function.

Thursday, 7 November 2013

$\displaystyle{y}=\sqrt{{{x}+\sqrt{{{x}+\sqrt{{{x}}}}}}}$ Find the derivative of the function.

$\displaystyle{y}=\sqrt{{{x}+\sqrt{{{x}+\sqrt{{{x}}}}}}}$

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

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

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

$\displaystyle{y}'={\left(\frac{{1}}{{{2}\sqrt{{{x}+\sqrt{{{x}+\sqrt{{{x}}}}}}}}}\right)}$
*(1+(1/(2sqrt(x+sqrt(x))))*(1+d/dxsqrt(x)))

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

$\displaystyle{y}'={\left(\frac{{1}}{{{2}\sqrt{{{x}+\sqrt{{{x}+\sqrt{{{x}}}}}}}}}\right)}\cdot{\left({1}+{\left(\frac{{1}}{{{2}\sqrt{{{x}+\sqrt{{{x}}}}}}}\right)}\right.}$
*((2sqrt(x)+1)/(2sqrt(x))))

$\displaystyle{y}'={\left(\frac{{1}}{{{2}\sqrt{{{x}+\sqrt{{{x}+\sqrt{{{x}}}}}}}}}\right)}$
*(1+(2sqrt(x)+1)/(4sqrt(x)sqrt(x+sqrt(x))))

$\displaystyle{y}'={\left(\frac{{1}}{{{2}\sqrt{{{x}+\sqrt{{{x}+\sqrt{{{x}}}}}}}}}\right)}$
*((4sqrt(x)sqrt(x+sqrt(x))+2sqrt(x)+1)/(4sqrt(x)sqrt(x+sqrt(x))))

$\displaystyle{y}'=\frac{{{4}\sqrt{{{x}}}\sqrt{{{x}+\sqrt{{{x}}}}}+{2}\sqrt{{{x}}}+{1}}}{{{8}\sqrt{{{x}}}\sqrt{{{x}+\sqrt{{{x}}}}}\sqrt{{{x}+\sqrt{{{x}+\sqrt{{{x}}}}}}}}}$

Notes

Thursday, 7 November 2013

$\displaystyle{y}=\sqrt{{{x}+\sqrt{{{x}+\sqrt{{{x}}}}}}}$ Find the derivative of the function.

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