3. ENG: What is the derivative of \( \frac{3 x^{2}}{1-\sin x} \) ?: FRE:Quel

To find the derivative of the function \( f(x) = \frac{3x^2}{1 - \sin x} \), we will use the quotient rule. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then the derivative is given by: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] In this case, let: - \( u = 3x^2 \) - \( v = 1 - \sin x \) Now, we need to find the derivatives \( u' \) and \( v' \): 1. \( u' = \frac{d}{dx}(3x^2) = 6x \) 2. \( v' = \frac{d}{dx}(1 - \sin x) = -\cos x \) Now we can apply the quotient rule: \[ f'(x) = \frac{u'v - uv'}{v^2} = \frac{(6x)(1 - \sin x) - (3x^2)(-\cos x)}{(1 - \sin x)^2} \] Now, simplify the numerator: \[ = \frac{6x(1 - \sin x) + 3x^2 \cos x}{(1 - \sin x)^2} \] Thus, the derivative of \( \frac{3x^2}{1 - \sin x} \) is: \[ f'(x) = \frac{6x(1 - \sin x) + 3x^2 \cos x}{(1 - \sin x)^2} \]