What is derivative of √arcsinβ ?

1 Answer
Jim G.
Mar 4, 2018

#1/(2sqrt(sin^-1beta(1-beta^2)))#

Explanation:

#"differentiate using the "color(blue)"chain rule"#

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#"note that "arcsinbeta=sin^-1beta"#

#"and that "d/dx(sin^-1x)=1/sqrt(1-x^2)#

#rArry=sqrt(sin^-1beta)=(sin^-1beta)^(1/2)#

#rArrdy/(dbeta)=1/2(sin^-1beta)^(-1/2)xx1/sqrt(1-beta^2)#

#color(white)(xxxxx)=1/(2sqrt(sin^-1beta)sqrt(1-beta^2))#

#color(white)(xxxxx)=1/(2sqrt(sin^-1beta(1-beta^2))#

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