How do you differentiate # y = (5x^2) / (4x-1)#? Calculus Basic Differentiation Rules Quotient Rule 1 Answer Himanshu Shekhar Jun 9, 2016 # dy/dx = (10x(2x-1))/ (4x-1)^2 # Just use the quotient rule and simplify. Explanation: # dy/dx = ((4x-1)* d/dx [ 5x^2 ] - 5x^2 * d/dx [ 4x-1 ] )/ (4x-1)^2 # # dy/dx = ((4x-1)* 10x - 5x^2 * 4 )/ (16x^2+1-8x) # # dy/dx = ((40x^2-10x) - 20x^2 )/ (16x^2+1-8x) # # dy/dx = (10x(2x-1))/ (4x-1)^2 # 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 4093 views around the world You can reuse this answer Creative Commons License