Find the derivative of the function \( y=x^{3} \cos ^{-1} x \) \( \begin{array}{l}\text { a) } \\ y \\ \prime \\ =3 x^{2} \cdot \frac{1}{\sqrt{1-x^{2}}}+\cos ^{-1}(x) \\ y^{\prime}=3 x^{2} \cos ^{-1}(x)+3 x^{2} \frac{(-1)}{\sqrt{1-x^{2}}} \\ \text { c) } y \prime=3 x^{2} \cos ^{-1}(x) \\ \text { d) } y \prime=\frac{-x^{3}}{\sqrt{1-x^{2}}}+3 x^{2} \cos ^{-1}(x)\end{array} \)

To find the derivative \( y' \) of the function \( y=x^{3} \cos ^{-1} x \), we will use the product rule, which states that if you have two functions \( u \) and \( v \), then \( (uv)' = u'v + uv' \). Here, let: \( u = x^{3} \) and \( v = \cos^{-1} x \). Now, calculate the derivatives: - \( u' = 3x^{2} \) - \( v' = \frac{-1}{\sqrt{1-x^{2}}} \) (the derivative of \( \cos^{-1} x \)) Applying the product rule: \[ y' = u'v + uv' = (3x^{2})(\cos^{-1} x) + (x^{3})\left(\frac{-1}{\sqrt{1-x^{2}}}\right) \] So, the final derivative is: \[ y' = 3x^{2} \cos^{-1} x - \frac{x^{3}}{\sqrt{1-x^{2}}} \] This matches the option: \( d) \ y' = 3 x^{2} \cos^{-1}(x) - \frac{x^{3}}{\sqrt{1-x^{2}}} \) (essentially option d but note the sign with the cosine inverse formulation). Make sure to double-check the options for any minor discrepancies!