What is the derivative of #sin^-1 * (5x)#?

Question

3642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-03T13:10:25

#(dy)/(dx)=5/(sqrt(1-25x^2))#

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

#=>5x=siny#

differentiate#" " wrt" " y#

#5(dx)/(dy)=cosy#

#(dx)/(dy)=cosy/5#

#=>(dy)/(dx)=5/cosy#

we need to rewrite this in terms of the original variable #x#

now:

#sin^2y+cos^2y=1#

#=>cosy=sqrt(1-sin^2y)#

#=>cosy=sqrt(1-(5x)^2)#

#:.cosy=sqrt(1-25x^2)#

#(dy)/(dx)=5/(sqrt(1-25x^2))#