What is the derivative of #f (x) = 3 arcsin (x^4)#?

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Answer 1 · 2016-07-09T04:56:42

# f'(x)=(12x^3)/{sqrt(1-x^8)}.#

Let #y=f(x)=3arcsinx^4.#

We know that #d/dt(arcsint)=1/sqrt(1-t^2).#

Let, #y=3arcsint#, where, #t=x^4.# Thus, #y# becomes a fun. of #t#, and, #t# of #x#.

Therefore, by Chain Rule, #dy/dx=dy/dt*dt/dx.#

#=d/dt(3arcsint)*d/dx(x^4),#

#=3*d/dt(arcsint)*4*x^(4-1),#
#=12*1/sqrt(1-t^2)*x^3,#
#=(12x^3)/[sqrt{1-(x^4)^2)},#
#:. f'(x)=(12x^3)/{sqrt(1-x^8)}.#

Enjoy Maths.!