What is the derivative of this function #y=sec^-1(5x)#?

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Answer 1 · 2018-05-05T12:49:51

#(dy)/(dx)=1/(|x|sqrt(25x^2-1))#

We know that,

#color(red)(d/(dt)(sec^-1t)=1/(|t|sqrt(t^2-1))#

Here,

#y=sec^-1(5x)#

Let.

#y=sec^-1u ,where, u=5x#

#=>(dy)/(du)=1/(|u|sqrt(u^2-1)) and (du)/(dx)=5#

#"Using "color(blue)"Integration by Parts"#

#color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)#

So,#(dy)/(dx)=1/(|u|sqrt(u^2-1))xx5#
#(dy)/(dx)=5/(|5x|sqrt(25x^2-1))#

#(dy)/(dx)=1/(|x|sqrt(25x^2-1))#