How do you differentiate # g(x) =cscx -x secx #? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer Bdub Apr 13, 2016 #g'(x)=-cscxcotx-xsecxtanx-secx# Explanation: #g'(x)=-cscxcotx-[xsecxtanx+secx]->#product rule #g'(x)=-cscxcotx-xsecxtanx-secx# 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 1706 views around the world You can reuse this answer Creative Commons License