How do you find the derivative of function #y=(sqrtx) *(x-3)^2#? Calculus Basic Differentiation Rules Product Rule 1 Answer Massimiliano Apr 20, 2015 In this way: #y'=1/(2sqrtx)*(x-3)^2+sqrtx*2(x-3)^(2-1)*1=# #=(x-3)[(x-3)/(2sqrtx)+2sqrtx]=(x-3)((x-3+4x)/(2sqrtx))=# #=((x-3)(5x-3))/(2sqrtx)#. 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 1666 views around the world You can reuse this answer Creative Commons License