Dy/dx? Ln(sin^-1(x))

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Answer 1 · 2018-04-24T18:20:12

#d/dxln(sin^-1x)=1/(sin^-1xsqrt(1-x^2))#

The Chain Rule, when applied to logarithms, tells us that if #u# is some function in terms of #x#, then

#d/dx(lnu)=1/u*(du/dx)#

For #ln(sin^-1x),# we see:

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

So,

#d/dxln(sin^-1x)=1/sin^-1x*1/sqrt(1-x^2)#

#d/dxln(sin^-1x)=1/(sin^-1xsqrt(1-x^2))#