How do you use the product rule to differentiate #g(s)=sqrts(4-s^2)#? Calculus Basic Differentiation Rules Product Rule 1 Answer Gerardina C. 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)# Answer link Related questions What is the Product Rule for derivatives? How do you apply the product rule repeatedly to find the derivative of #f(x) = (x - 3)(2 - 3x)(5 - x)# ? How do you use the product rule to find the derivative of #y=x^2*sin(x)# ? How do you use the product rule to differentiate #y=cos(x)*sin(x)# ? How do you apply the product rule repeatedly to find the derivative of #f(x) = (x^4 +x)*e^x*tan(x)# ? How do you use the product rule to find the derivative of #y=(x^3+2x)*e^x# ? How do you use the product rule to find the derivative of #y=sqrt(x)*cos(x)# ? How do you use the product rule to find the derivative of #y=(1/x^2-3/x^4)*(x+5x^3)# ? How do you use the product rule to find the derivative of #y=sqrt(x)*e^x# ? How do you use the product rule to find the derivative of #y=x*ln(x)# ? See all questions in Product Rule Impact of this question 1445 views around the world You can reuse this answer Creative Commons License