What is the derivative of $y = Arcsin ((3x)/4)$ ?

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Answer 1 · 2015-07-06T12:39:34

Use the Chain rule and the definition of the derivative of the arcsin of a function.

First, understand that $y=sin^-1(3x/4)$ is way of saying $y=f(g(x))$

Second, let $f(m) = sin^-1(m)$ (i.e., the "outside" part) and $m=g(x)=3x/4$ (i.e., the "inside" part).

Third, use the Chain Rule which states $dy/dx=dy/(dm)*(dm)/(dx)=f'(m)g'(x)$

Fourth, find $(dy)/(dm)$

$(dy)/(dm)= f'(m) = d/(dm)(sin^-1(m))=1/sqrt(1-x^2)$

Fifth, find $(dm)/(dx)$

$(dm)/(dx)=g'(x)=d/(dx)(3x/4)=3/4$

Finally, multiply your results:

$(dy)/(dx)=(dy)/(dm)(dm)/(dx)=1/sqrt(1-x^2)*3/4=3/(4sqrt(1-x^2))$ .

Done, and done!

What is the derivative of y = Arcsin ((3x)/4) y = Arcsin ((3x)/4)?