How do you differentiate #g(y) =(x^2 + 6)sqrtx # using the product rule? Calculus Basic Differentiation Rules Product Rule 1 Answer Monzur R. Jan 13, 2017 #g'y=2x^(3/2)+(x^2+6)/(2sqrt(x))# Explanation: #g(y)=(x^2+6)sqrt(x)=(x^2+6)(x^(1/2))# Product rule: #f'(uv)=vu'+uv'# #u=x^2+6, v=x^(1/2)# #u'=2x, v'=1/2x^(-1/2)=1/(2sqrt(x))# #g'y=2x^(3/2)+(x^2+6)/(2sqrt(x))# 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 1108 views around the world You can reuse this answer Creative Commons License