Find the derivative of f(x) = 1+tan(x) / 1-tan(x)

Tuesday, October 23, 2012

Find the derivative of f(x) = 1+tan(x) / 1-tan(x)

We need to know
the following trigonometric function here.

$\displaystyle{\cos{{2}}}{x}={{\cos}}^{{x}}-{{\sin}}^{{2}}{x}$

$\displaystyle{\sin{{2}}}{x}={2}{\sin{{x}}}{\cos{{x}}}$

$\displaystyle{{\sin}}^{{2}}{x}+{{\cos}}^{{2}}{x}={1}$

$\displaystyle\frac{{{1}+{\tan{{x}}}}}{{{1}-{\tan{{x}}}}}$

$\displaystyle=$
(1+tanx)/(1-tanx)xx(1+tanx)/(1+tanx)

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

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

$\displaystyle=$
((cosx+sinx)^2/(cos^2x))/((cos^2x-sin^2x)/(cos^2x))

$\displaystyle=$
(cosx+sinx)^2/(cos^2x-sin^2x)

$\displaystyle=$
(cos^2x+sin^2x+2sinxcosx)/(cos2x)

$\displaystyle=\frac{{{1}+{\sin{{2}}}{x}}}{{{\cos{{2}}}{x}}}$

$\displaystyle{f{{\left({x}\right)}}}=\frac{{{1}+{\tan{{x}}}}}{{{1}-{\tan{{x}}}}}=\frac{{{1}+{\sin{{2}}}{x}}}{{{\cos{{2}}}{x}}}$

$\displaystyle{f{'}}{\left({x}\right)}$

$\displaystyle=$
(cos2x*2cos2x-(1+sin2x)(-2sin2x))/(cos^2(2x))

$\displaystyle=$
(2(cos^2(2x)+sin^2(2x)+sin2x))/(cos^2(2x))

$\displaystyle=\frac{{{2}{\left({1}+{\sin{{2}}}{x}\right)}}}{{{{\cos}}^{{2}}{\left({2}{x}\right)}}}$

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

$\displaystyle=$
(2(1+sin2x))/((1+sin2x)(1-sin2x))

$\displaystyle=\frac{{2}}{{{1}-{\sin{{2}}}{x}}}$

So the derivative of f(x)
is $\displaystyle{f{'}}{\left({x}\right)}=\frac{{2}}{{{1}-{\sin{{2}}}{x}}}$

Spooky Notes

Tuesday, October 23, 2012

Find the derivative of f(x) = 1+tan(x) / 1-tan(x)

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