What is the derivative of #tan(5x)^(1/2)#?

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Answer 1 · 2016-10-27T08:13:09

#(dy)/(dx)=sqrt5/2(x^(-1/2))sec^2(5x)^(1/2)#

#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)#