What is the derivative of #x*sqrt(4-x)#? Calculus Basic Differentiation Rules Product Rule 1 Answer Narad T. Nov 3, 2016 The answer is #=(8-3x)/(2sqrt(4-x))# Explanation: Use #(uv)'=u'v+uv'# #u=x##=>##u'=1# #v=sqrt(4-x)##=>##v'=1/(2sqrt(4-x))*-1# #(x*sqrt(4-x))'=1*sqrt(4-x)-x/(2sqrt(4-x))# #=(2(4-x)-x)/(2sqrt(4-x))=(8-2x-x)/(2sqrt(4-x))=(8-3x)/(2sqrt(4-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 12506 views around the world You can reuse this answer Creative Commons License