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

1 Answer

Feb 2, 2017

#g'(s)=(4-5s^2)/(2sqrts)#

Explanation:

#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)#

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