Differentiating

[shaunjpeterson]

When differentiatiing y= 1- cos x / sinx you go sin x -cosx(sin x) / sin^2 x right? If im right what is my next step

 

[tkhunny]

I can't follow it. Notation needs work. Is it: $\displaystyle \frac{1-cos(x)}{sin(x)}$ or $\displaystyle 1-\frac{cos(x)}{sin(x)}$?

$\displaystyle \frac{d}{dx}(\frac{1-cos(x)}{sin(x)})\,=\,\frac{sin(x)*(sin(x))\,-\,(1-cos(x))*cos(x)}{[sin(x)]^{2}}$

 

[Euler]

I came up with a different answer:

You are trying to find the derivative of the quotient $\displaystyle \frac{1-cos(x)}{sin(x)}$

I am sure your book gave you an equation that tells you what to do:$\displaystyle \frac{u'v-uv'}{v^2}$

Remember that the derivative of a constant is zero.

$\displaystyle \frac{1-cos(x)}{sin(x)} \Rightarrow \frac{0-(-sin(x))(sin(x))-(1-cos(x))(cos(x))}{sin^2(x)}$

Darn, tk you beat me to it!