Question

differentiate the function with respect to the independent variable.
$f(x)=\frac{x^{3}}{15}-\frac{x^{4}}{12}+\frac{2}{15}$
$f(x)=\square$

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:

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