I need help finding the derivative of the following function:
y = sinx/1-secx *I know that the answer should be cosx-1+tan^2(x)/(1-secx)^2 but am having trouble working it out.
Quotient rule: $\displaystyle \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$
So: $\displaystyle \left(\frac{\sin x}{1 - \sec x}\right)' = \frac{(\sin x)'(1 - \sec x) - \sin x (1 - \sec x)'}{(1 - \sec x)^2}$
$\displaystyle = \frac{\cos x (1 - \sec x) - \sin x (-\sec x \tan x)}{(1-\sec x)^2}$
And simplify ...