differentiate the function with respect to the independent variable.\n$f(x)=\\frac{x^{3}}{15}-\\frac{x^{4}}{1…

Explanation:

Step1: Apply power rule

The power rule states that if (y = x^n), then (y^\prime=nx^{n - 1}). For the function (f(x)=\frac{x^{3}}{15}-\frac{x^{4}}{12}+\frac{15}{2}), we differentiate each term separately. For the first term (\frac{x^{3}}{15}), using the power rule: (\frac{d}{dx}(\frac{x^{3}}{15})=\frac{1}{15}\times3x^{3 - 1}=\frac{3x^{2}}{15}=\frac{x^{2}}{5}). For the second term (-\frac{x^{4}}{12}), using the power rule: (\frac{d}{dx}(-\frac{x^{4}}{12})=-\frac{1}{12}\times4x^{4 - 1}=-\frac{4x^{3}}{12}=-\frac{x^{3}}{3}). For the third term (\frac{15}{2}) (a constant), since the derivative of a constant (C) is (0), (\frac{d}{dx}(\frac{15}{2}) = 0).

Step2: Combine the derivatives of each term

(f^\prime(x)=\frac{d}{dx}(\frac{x^{3}}{15}-\frac{x^{4}}{12}+\frac{15}{2})=\frac{d}{dx}(\frac{x^{3}}{15})-\frac{d}{dx}(\frac{x^{4}}{12})+\frac{d}{dx}(\frac{15}{2})) Substitute the derivatives of each term: (f^\prime(x)=\frac{x^{2}}{5}-\frac{x^{3}}{3}+0)

Answer:

(\frac{x^{2}}{5}-\frac{x^{3}}{3})