What is the derivative of #tan(5x)^(1/2)#? Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 1 Answer sjc Oct 27, 2016 #(dy)/(dx)=sqrt5/2(x^(-1/2))sec^2(5x)^(1/2)# Explanation: #y=tan(5x)^(1/2)# #u=(5x)^(1/2)=5^(1/2)x^(1/2)# #=>(dy)/(du)=5^(1/2)1/2x^(-1/2)=sqrt5/2(x^(-1/2))# #y=tanu=>(dy)/(du)=sec^2u# #(dy)/(dx)=(dy)/(du)xx(du)/(dx)# #:.(dy)/(dx)=sec^2uxxsqrt5/2(x^(-1/2))# #:.(dy)/(dx)=sqrt5/2(x^(-1/2))sec^2(5x)^(1/2)# Answer link Related questions What is the derivative of #y=cos(x)# ? What is the derivative of #y=tan(x)# ? How do you find the 108th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x)# from first principle? How do you find the derivative of #y=cos(x^2)# ? How do you find the derivative of #y=e^x cos(x)# ? How do you find the derivative of #y=x^cos(x)#? How do you find the second derivative of #y=cos(x^2)# ? How do you find the 50th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x^2)# ? See all questions in Derivative Rules for y=cos(x) and y=tan(x) Impact of this question 2145 views around the world You can reuse this answer Creative Commons License