How do you differentiate #arcsin(csc(4x)) )# using the chain rule?

Question

1696 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-17T16:44:27

#d/dx(sin^-1 csc (4x))=4*sec 4x*sqrt(1-csc^2 4x)#

We use the formula

#d/dx(sin^-1 u)=(1/sqrt(1-u^2))du#

#d/dx(sin^-1 csc (4x))=(1/sqrt(1-(csc 4x)^2))d/dx(csc 4x)#

#d/dx(sin^-1 csc (4x))=(1/sqrt(1-csc^2 4x))*(-csc 4x*cot 4x)*d/dx(4x)#

#d/dx(sin^-1 csc (4x))=((-csc 4x*cot 4x)/sqrt(1-csc^2 4x))*(4)#

#d/dx(sin^-1 csc (4x))=((-4*csc 4x*cot 4x)/sqrt(1-csc^2 4x))*(sqrt(1-csc^2 4x)/(sqrt(1-csc^2 4x)))#

#d/dx(sin^-1 csc (4x))=((-4*csc 4x*cot 4x*sqrt(1-csc^2 4x))/(-cot^2 4x))#

#d/dx(sin^-1 csc (4x))=4*sec 4x*sqrt(1-csc^2 4x)#

God bless....I hope the explanation is useful.