If y=ln(2x^2-6x), then what is dy/dx? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Narad T. Dec 19, 2016 The answer is =(2x-3)/(x(x-3)) Explanation: This is a chain differentiation y=ln u(x) dy/dx=1/(u(x))*u'(x) y=ln(2x^2-6x) dy/dx=1/(2x^2-6x)*(4x-6) =(2x-3)/(x^2-3x)=(2x-3)/(x(x-3)) Answer link Related questions How do you solve separable differential equations? How do you solve separable first-order differential equations? How do you solve separable differential equations with initial conditions? What are separable differential equations? How do you solve the differential equation dy/dx=6y^2x, where y(1)=1/25 ? How do you solve the differential equation y'=e^(-y)(2x-4), where y5)=0 ? How do you solve the differential equation (dy)/dx=e^(y-x)sec(y)(1+x^2), where y(0)=0 ? How do I solve the equation dy/dt = 2y - 10? Given the general solution to t^2y'' - 4ty' + 4y = 0 is y= c_1t + c_2t^4, how do I solve the... How do I solve the differential equation xy'-y=3xy, y_1=0? See all questions in Solving Separable Differential Equations Impact of this question 3754 views around the world You can reuse this answer Creative Commons License