Question

find the indicated derivative.

\\( \frac { d r } { d s } \\) if \\( r = s ^ { 3 } - 4 s ^ { 2 } + 1 \\)

\\( \frac { d r } { d s } = \\)

Explanation:

Step1: Differentiate each term

Differentiate \(s^3\) using the power rule \(\frac{d}{ds}(s^n)=ns^{n - 1}\), so \(\frac{d}{ds}(s^3)=3s^{2}\).
Differentiate \(-4s^2\) using the power rule, \(\frac{d}{ds}(-4s^2)=-4\times2s=-8s\).
Differentiate the constant term \(1\), \(\frac{d}{ds}(1) = 0\).

Step2: Combine the derivatives

By the sum - rule of differentiation \(\frac{d}{ds}(u + v+w)=\frac{du}{ds}+\frac{dv}{ds}+\frac{dw}{ds}\) (where \(u = s^3\), \(v=-4s^2\), \(w = 1\)), we have \(\frac{dr}{ds}=\frac{d}{ds}(s^3-4s^2 + 1)=\frac{d}{ds}(s^3)-\frac{d}{ds}(4s^2)+\frac{d}{ds}(1)\).

Answer:

\(3s^{2}-8s\)