What is the derivative of #(6)/(x^3sqrtx)#? Calculus Basic Differentiation Rules Quotient Rule 1 Answer Massimiliano Apr 30, 2015 The answer is: #y'=-21/(x^4sqrtx)#. The function can be written: #y=6*x^(-3-1/2)rArry=6x^(-7/2)# #y'=6*(-7/2)*x^(-7/2-1)=-21x^(-9/2)=-21/sqrt(x^9)=# #=-21/(x^4sqrtx)#. Answer link Related questions What is the Quotient Rule for derivatives? How do I use the quotient rule to find the derivative? How do you prove the quotient rule? How do you use the quotient rule to differentiate #y=(2x^4-3x)/(4x-1)#? How do you use the quotient rule to differentiate #y=cos(x)/ln(x)#? How do you use the quotient rule to find the derivative of #y=tan(x)# ? How do you use the quotient rule to find the derivative of #y=x/(x^2+1)# ? How do you use the quotient rule to find the derivative of #y=(e^x+1)/(e^x-1)# ? How do you use the quotient rule to find the derivative of #y=(x-sqrt(x))/(x^(1/3))# ? How do you use the quotient rule to find the derivative of #y=x/(3+e^x)# ? See all questions in Quotient Rule Impact of this question 1329 views around the world You can reuse this answer Creative Commons License