if y= log{cot(pi/4+x/2)}. find dy/dx

Wednesday, October 12, 2011

if y= log{cot(pi/4+x/2)}. find dy/dx

$\displaystyle{y}={\log{{\left\lbrace{\cot{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}\right\rbrace}}}$

Let $\displaystyle{u}={\cot{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}$

Apply the chain rule,

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{d}}{{{d}{u}}}{\left({\log}_{{10}}{\left({u}\right)}\right)}\frac{{d}}{{\left.{d}{x}}}{\left({\cot{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}\right)}$

Now apply the common derivative,

$\displaystyle\frac{{d}}{{{d}{u}}}{\left({\log}_{{10}}{\left({u}\right)}\right)}=\frac{{1}}{{{u}{\ln{{\left({10}\right)}}}}}$

Now let's evaluate $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({\cot{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}\right)}$ by chain rule,

Let $\displaystyle{u}=\frac{\pi}{{4}}+\frac{{x}}{{2}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({\cot{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}\right)}=\frac{{d}}{{{d}{u}}}{\left({\cot{{\left({u}\right)}}}\right)}\frac{{d}}{{\left.{d}{x}}}{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}$

Use the common derivative $\displaystyle\frac{{d}}{{{d}{u}}}{\left({\cot{{\left({u}\right)}}}\right)}=-{{\csc}}^{{2}}{\left({u}\right)}$

and $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}=\frac{{1}}{{2}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({\cot{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}\right)}=-{{\csc}}^{{2}}{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}\cdot\frac{{1}}{{2}}$

and $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={\left(\frac{{1}}{{{u}{\ln{{\left({10}\right)}}}}}\right)}{\left(-\frac{{1}}{{2}}{{\csc}}^{{2}}{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}\right)}$

Substitute back $\displaystyle{u}={\cot{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={\left(\frac{{1}}{{{\cot{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}{\ln{{\left({10}\right)}}}}}\right)}{\left(-\frac{{1}}{{2}}{{\csc}}^{{2}}{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}\right.}$

$\displaystyle=-\frac{{1}}{{{2}{\ln{{\left({10}\right)}}}}}\frac{{{{\csc}}^{{2}}{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}{{\cot{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}}$

We can further simplify as $\displaystyle{{\csc}}^{{2}}{\left(\theta\right)}=\frac{{1}}{{{{\sin}}^{{2}}{\left(\theta\right)}}}$

and $\displaystyle{\cot{{\left(\theta\right)}}}=\frac{{\cos{{\left(\theta\right)}}}}{{{\sin{{\left(\theta\right)}}}}}$

$\displaystyle\therefore\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=-\frac{{1}}{{{2}{\ln{{\left({10}\right)}}}}}\frac{{1}}{{{{\sin}}^{{2}}{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}{\left(\frac{{\sin{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}}{{{\cos{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}}}\right)}$

$\displaystyle=-\frac{{1}}{{{2}{\ln{{\left({10}\right)}}}{\sin{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}{\cos{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}}}$

This can be simplified combining terms.

$\displaystyle-\frac{{1}}{{{\ln{{\left({10}\right)}}}{2}{\sin{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}{\cos{{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}}}}}$

$\displaystyle=-\frac{{1}}{{{\ln{{\left({10}\right)}}}{\sin{{\left({2}{\left(\frac{\pi}{{4}}+\frac{{x}}{{2}}\right)}\right)}}}}}$

$\displaystyle=-\frac{{1}}{{{\ln{{\left({10}\right)}}}{\sin{{\left(\frac{\pi}{{2}}+{x}\right)}}}}}=-\frac{{1}}{{{\ln{{\left({10}\right)}}}{\cos{{\left({x}\right)}}}}}$

This is the answer: $\displaystyle\frac{{-{\sec{{\left({x}\right)}}}}}{{\ln{{\left({10}\right)}}}}$

Ninenotes

Wednesday, October 12, 2011

if y= log{cot(pi/4+x/2)}. find dy/dx

No comments:

Post a Comment

find square roots of -1+2i