How do you find the derivative of # y =cotx/(1-sinx)#?

Question

Ezekiel Alexander · Accepted Answer

Use the quotient rule:

If we let #cotx/(1-sinx)=f(x)/g(x)#

Then, to find the derivative using the quotient rule would generally look like:

#(f'(x)*g(x)-g'(x)*f(x))/g(x)^2#

Therefore, #y'# would look like:

#y'=[d/dxcotx*(1-sinx)-d/dx(1-sinx)*cotx]/(1-sinx)^2#

Find the derivatives and simplify.

Adam Adkins · Answer

undefined To find the derivative of $ y = \frac{\cot x}{1 - \sin x} $, you can use the quotient rule.

1. Identify $ u(x) = \cot x $ and $ v(x) = 1 - \sin x $.
2. Apply the quotient rule, which states that the derivative of $ \frac{u(x)}{v(x)} $ is given by:
   $$ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$
3. Differentiate $ u(x) $ and $ v(x) $ separately:
   $$ u'(x) = -\csc^2 x $$
   $$ v'(x) = -\cos x $$
4. Substitute the derivatives and the original functions into the quotient rule formula:
   $$ y' = \frac{(-\csc^2 x)(1 - \sin x) - (\cot x)(-\cos x)}{(1 - \sin x)^2} $$
5. Simplify the expression:
   $$ y' = \frac{-\csc^2 x + \csc^2 x \sin x + \cot x \cos x}{(1 - \sin x)^2} $$
   $$ y' = \frac{\csc^2 x(\sin x - 1) + \cot x \cos x}{(1 - \sin x)^2} $$

This is the derivative of $ y = \frac{\cot x}{1 - \sin x} $ using the quotient rule.