Differentiate the function \( B(x)=\frac{\frac{3}{\sin x}-3 \sin x}{4 \cos x} \)

To differentiate the function \( B(x)=\frac{\frac{3}{\sin x}-3 \sin x}{4 \cos x} \), we apply the quotient rule. The quotient rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] Here, let: - \( u = \frac{3}{\sin x} - 3 \sin x \) - \( v = 4 \cos x \) First, we need to differentiate \( u \) and \( v \). 1. **Finding \( u' \)**: To differentiate \( u \), we use the derivative rules for \( \frac{3}{\sin x} \) and \( -3 \sin x \): \[ u' = \frac{d}{dx}\left(\frac{3}{\sin x}\right) - 3 \frac{d}{dx}(\sin x) \] \[ = -\frac{3 \cos x}{\sin^2 x} - 3 \cos x \] \[ = -\frac{3 \cos x + 3 \cos x \sin^2 x}{\sin^2 x} \] \[ = -\frac{3 \cos x(1 + \sin^2 x)}{\sin^2 x} \] 2. **Finding \( v' \)**: \[ v' = \frac{d}{dx}(4 \cos x) = -4 \sin x \] 3. **Applying the Quotient Rule**: Now substituting \( u, v, u', v' \) into the quotient formula: \[ B'(x) = \frac{(-\frac{3 \cos x(1 + \sin^2 x)}{\sin^2 x})(4 \cos x) - \left(\frac{3}{\sin x} - 3 \sin x\right)(-4 \sin x)}{(4 \cos x)^2} \] Simplifying the numerator: Calculating the first term: \[ -\frac{3 \cos x(1 + \sin^2 x)}{\sin^2 x} \cdot 4 \cos x = -\frac{12 \cos^2 x(1 + \sin^2 x)}{\sin^2 x} \] Calculating the second term: \[ -4 \sin x \left(\frac{3}{\sin x} - 3 \sin x\right) = -4 \left(3 - 3 \sin^2 x\right) = -12 + 12 \sin^2 x \] Putting it all together, the full derivative is: \[ B'(x) = \frac{-\frac{12 \cos^2 x(1 + \sin^2 x)}{\sin^2 x} + (-12 + 12 \sin^2 x)}{16 \cos^2 x} \] Now you have a greasy mix of trigonometric derivatives, and simplifying that should lead to your final result! Do take care with the signs and combine like terms judiciously. Happy differentiating!