shaunjpeterson
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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
Differentiating
- Thread starter shaunjpeterson
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I can't follow it. Notation needs work. Is it: \(\displaystyle \frac{1-cos(x)}{sin(x)}\) or \(\displaystyle 1-\frac{cos(x)}{sin(x)}\)?
\(\displaystyle \frac{d}{dx}(\frac{1-cos(x)}{sin(x)})\,=\,\frac{sin(x)*(sin(x))\,-\,(1-cos(x))*cos(x)}{[sin(x)]^{2}}\)
\(\displaystyle \frac{d}{dx}(\frac{1-cos(x)}{sin(x)})\,=\,\frac{sin(x)*(sin(x))\,-\,(1-cos(x))*cos(x)}{[sin(x)]^{2}}\)
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!
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!