How do you differentiate # g(x) = sqrt(arc csc(x^3+1) #?

1 Answer
Guilherme N.
Jan 4, 2016

Supposing #|(x^3+1)|>1#, we can proceed using the specific rule for #arccsc(f(x))# and also chain rule for the square root.

Explanation:

  • #(arccscu)'=(-u')/(|u|sqrt(u^2-1))#

  • Chain rule: #(dy)/(dx)=(dy)/(dv)(dv)/(dx)#

Renaming #v=arccsc(x^3+1)#, we have #g(x)=sqrt(v)#

#(dg(x))/(dx)=1/(2v)((-3x^2)/(|x^3+1|sqrt(x^3(x^3+2))))=((-3x^2)/(arccsc(x^3+1)|x^3+1|sqrt(x^3(x^3+2))))#

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