How do you differentiate # g(x) =4/x-cscx #? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer Trevor Ryan. Jun 10, 2016 #g'(x)=(4/x-cosecx)=-4/x^2+cosecx*cotx# Explanation: #d/dx(4/x-cosecx)=-4/x^2+cosecx*cotx# Answer link Related questions What is Derivatives of #y=sec(x)# ? What is the Derivative of #y=sec(x^2)#? What is the Derivative of #y=x sec(kx)#? What is the Derivative of #y=sec ^ 2(x)#? What is the derivative of #y=4 sec ^2(x)#? What is the derivative of #y=ln(sec(x)+tan(x))#? What is the derivative of #y=sec^2(x)#? What is the derivative of #y=sec^2(x) + tan^2(x)#? What is the derivative of #y=sec^3(x)#? What is the derivative of #y=sec(x) tan(x)#? See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x) Impact of this question 1620 views around the world You can reuse this answer Creative Commons License