How do you differentiate #f(x)=2cscx+5cosx#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer Tazwar Sikder May 20, 2017 #f'(x) = - 2 csc(x) cot(x) - 5 sin(x)# Explanation: We have: #f(x) = 2 csc(x) + 5 cos(x)# #Rightarrow f'(x) = frac(d)(dx)(2 csc(x)) + frac(d)(dx)(5 cos(x))# #Rightarrow f'(x) = 2 (- csc(x) cot(x)) + 5 (- sin(x))# #Rightarrow f'(x) = - 2 csc(x) cot(x) - 5 sin(x)# 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 3931 views around the world You can reuse this answer Creative Commons License