Derivative Trig Help !!!

[DancingQueen09]

I look at this problem and I dont even know how to start... even worse complete it!!! If someone could please show me step by step how to get the answer to this problem I would greatly appreciate it : )

If

Find f'(x) (the first derivative)

Thank you in advance <3<3<3

 

[galactus]

One little thing you could do is rewrite $\displaystyle ln(\sqrt{1-cos(x)})=\frac{1}{2}ln(1-cos(x))$

Use the chain rule. It is not as bad as it looks.

i.e. $\displaystyle \frac{d}{dx}[sin(ln(x))]=\frac{cos(ln(x))}{x}$

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ \frac{x}{2}[sin(ln|x|)+cos(ln|x|)]+ln(1-cos|x|)^{1/2}, \ find \ f \ ' \ (x).$

$\displaystyle Now. instead \ of \ having \ a \ conniption \ fit \ or \ leaving \ for \ the \ nearest \ trauma \ center, \ break$

$\displaystyle the \ equation \ down \ to \ parts \ that \ are \ more \ manageable, \ to \ wit:$

$\displaystyle \frac{x}{2}[sin(ln|x|)+cos(ln|x|)]+\frac{1}{2}ln(1-cos|x|)$

$\displaystyle = \ \frac{1}{2}[xsin(ln|x|)+xcos(ln|x|)+ln(1-cos|x|)].$

$\displaystyle = \ \frac{1}{2}\bigg[(1)sin(ln|x|)+\frac{xcos(ln|x|)}{x}+(1)cos(ln|x|)-\frac{xsin(ln|x|)}{x}+\frac{sin(x)}{1-cos(x)}\bigg]$

$\displaystyle = \ \frac{1}{2}\bigg[sin(ln|x|)+cos(ln|x|)+cos(ln|x|)-sin(ln|x|)+\frac{sin(x)}{1-cos(x)}\bigg]$

$\displaystyle = \ cos(ln|x|)+\bigg(\frac{1}{2}\bigg)\frac{sin(x)}{1-cos(x)} \ QED$

$\displaystyle Now, \ all \ in \ all, \ that \ wasn't \ so \ bad, \ now \ was \ it.$