How do you use the product rule to differentiate #g(s)=sqrts(4-s^2)#?

Question

Elijah Abram · Accepted Answer

#g'(s)=(4-5s^2)/(2sqrts)# #g'(s)=1/(2sqrts)* (4 - s^2)+sqrts*(-2s)# #=(4-s^2)/(2sqrts)-2ssqrts# #(4-s^2-2ssqrts*(2sqrts))/(2sqrts)# #(4-s^2-4s^2)/(2sqrts)# #(4-5s^2)/(2sqrts)#

Emilia Aguilar · Answer

undefined To differentiate $ g(s) = \sqrt{4 - s^2} $ using the product rule, you first recognize it as the product of two functions: $ f(s) = \sqrt{4} $ and $ h(s) = \sqrt{4 - s^2} $. Then, apply the product rule $ (f \cdot h)' = f' \cdot h + f \cdot h' $ where $ f' $ is the derivative of $ f $ and $ h' $ is the derivative of $ h $. Differentiate each function separately and then apply the product rule.

$ f(s) = \sqrt{4} $ is a constant function, so its derivative $ f'(s) $ is zero.

$ h(s) = \sqrt{4 - s^2} $ can be differentiated using the chain rule: $ h'(s) = \frac{-s}{\sqrt{4 - s^2}} $.

Now, apply the product rule: 
$$ g'(s) = f'(s) \cdot h(s) + f(s) \cdot h'(s) $$
$$ g'(s) = 0 \cdot \sqrt{4 - s^2} + \sqrt{4} \cdot \frac{-s}{\sqrt{4 - s^2}} $$

Simplify:
$$ g'(s) = \frac{-s\sqrt{4}}{\sqrt{4 - s^2}} $$
$$ g'(s) = \frac{-s \cdot 2}{\sqrt{4 - s^2}} $$
$$ g'(s) = \frac{-2s}{\sqrt{4 - s^2}} $$