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Whats the derivative of #sin^-1 * (5x)#?

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Konstantinos Michailidis

Sep 27, 2015

See explanation

Explanation:

Set #arcsinx=cos^(-1)x# hence we know that

#d((arcsinx))/dx=1/(sqrt(1-x^2)#

hence for a function #u(x)#

#d(arcsinu(x))/dx=1/sqrt(1-u^2(x))*du/dx#

hence for #u=5x# we get

#d(arcsin5x)/dx=5/(sqrt(1-25x^2))#

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