Differentiating $f(x)=\frac{3(1-\sin x)}{2\cos x}$

Question

I'm having trouble to solve the following derivative. I tried to apply the quotient rule, but I cannot get the same result as my textbook.

The function:

$$f(x)=\frac{3(1-\sin x)}{2\cos x}$$

The result I supposed to get:

$$f'(x)=\frac{3}{2}\sec x(\tan x-\sec x)$$

I have tried this by applying the quotient rule:

$$f(x) =\frac{3(1-\sin x)}{2\cos x}=\frac{3-3\sin x}{2\cos x}$$ \begin{align*} f'(x) & =\frac{-3\cos x \cdot 2\cos x - (3- \sin x) \cdot -2\sin x}{2\cos^2x}\\ & =\frac{-6\cos^2x-(-6\sin x+6\sin^2x)}{2\cos^2x}\\ & =\frac{-6\cos^2x+6\sin x+6\sin^2x}{2\cos^2x} \end{align*}

Is someone able to show me step by step how to get the above result?

Thank you in advance for all of your help!

Answer 1

$$f(x)=\frac{3(1-\sin x)}{2\cos x} =\frac {3}{2} (\frac {1-\sin x}{\cos x})$$

Apply the quotient rule $$ f'(x) =\frac {3}{2}( \frac {-\cos x (\cos x)+\sin x (1-\sin x)}{\cos ^2 x})=$$

$$\frac {3}{2}( \frac {\sin x -1}{\cos ^2 x})=$$

$$\frac {3}{2}( \frac {1}{\cos x} \frac {\sin x -1}{\cos x})=$$

$$\frac {3}{2} \sec x(\tan x - sec x)$$