What is the derivative of #y=arcsin(x/7)#?

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Answer 1 · 2018-04-16T15:23:50

#d/dx arcsin(x/7)=(\sqrt{49-x^2})^{-1}#

We want to evaluate #d/dx arcsin(x/7)#.

We can apply the chain rule, that #(f@g)'(x)=(f'@g)(x)g'(x)#, and in our case #f=arcsin# and #g(x)=1/7x#. Therefore:

#d/dx arcsin(x/7)=\frac{1}{\sqrt{1-x^2/7^2}}\cdot d/dx x/7#

#d/dx arcsin(x/7)=\frac{1/7 d/dx x}{\sqrt{1-x^2/7^2}}#

#d/dx arcsin(x/7)=\frac{1}{7\sqrt{1-x^2/49}}#

We can further simplify this:

#d/dx arcsin(x/7)=\frac{1}{\sqrt{49(1-x^2/49)}}#

#d/dx arcsin(x/7)=(\sqrt{49-x^2})^{-1}#