Question

Use the Product Rule to differentiate the function. $$ g(s)=\sqrt{s}\left(4-s^{2}\right) $$

Short answer

The derivative of the function \( g(s)=\sqrt{s}\times(4-s^{2}) \) is \( g'(s) = 2*\sqrt{s} - 3s^{1.5} \).

Step-by-step solution

State the Product Rule

The Product Rule for differentiation is \( (f \cdot g)' = f' \cdot g + f \cdot g' \) i.e., the derivative of the product of two functions (f and g here) is equal to derivative of the first function times the second function plus the first function times the derivative of the second function.

Identify the Functions for Product Rule

In our case, the function \( g(s)=\sqrt{s}\times(4-s^{2}) \) can be broken down into two functions for the application of product rule. Consider \( f(s)=\sqrt{s} \) and \( g(s)=(4-s^{2}) \). The requirement now is to find \( f'(s) \) and \( g'(s) \).

Calculate the Derivatives

Differentiate each of \( f(s) \) and \( g(s) \). For \( f(s)=\sqrt{s} = s^{1/2} \), the derivative \( f'(s) \) is \( 0.5*(s^{-0.5}) = 1/(2*\sqrt{s}) \). For \( g(s)=(4-s^{2}) \), the derivative \( g'(s) \) is \( 0-2s = -2s \).

Apply the Product Rule

Given \( g'(s) = f'(s)\cdot g(s) + f(s)\cdot g'(s) \), substituting \( f'(s)= 1/(2*\sqrt{s}) \), \( g'(s) = -2s \), \( f(s)=\sqrt{s} = s^{1/2} \), and \( g(s)=(4-s^{2}) \), we find \( g'(s) = (1/(2*\sqrt{s}))* (4-s^{2}) + \sqrt{s}*(-2s) \).

Simplify the Expression

Finally, simplifying we get \( g'(s) = 2*\sqrt{s} - 3s^{1.5} \).