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

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Answer 1 · 2016-01-04T23:12:10

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

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

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

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