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diff. inverse trig (1 Viewer)

Thread starter eunkuk91
Start date Oct 27, 2009

eunkuk91

Member

Oct 27, 2009

can someone tell me what happens if its
sin^-1(1/f(x)) and cos^-1(1/f(x))

$y=sin^{-1}\frac{1}{f(x)}=sin^{-1}u, \text{ where }u=\frac{1}{f(x)}\\\frac{dy}{du}=\frac{1}{\sqrt{1-u^2}}=\frac{1}{\sqrt{1-(\frac{1}{f(x)})^2}}\\\frac{du}{dx}=-\frac{1}{[f(x)]^2}\\\therefore \text{By the chain rule }\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=-\frac{1}{[f(x)]^2}\times\frac{1}{\sqrt{1-(\frac{1}{f(x)})^2}}$

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