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find dy/dx using implicit differentiation using cosy=x for y=cos^(1)x
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\[ \cos(y) = x \\ \text{Differentiate implicitly with respect to x:} \\ -\sin(y)y' = 1 \\ y' = -\frac{1}{\sin(y)} \\ \sin(y) = \sqrt{1-\cos^2(y)} = \sqrt{1-x^2} \\ y' = -\frac{1}{\sqrt{1-x^2}} \]
The question is dy/dx(cos^(-1)x
\[ y = \cos^{-1}(x) \\ \frac{dy}{dx} = ? \\ \text{If }y = \cos^{-1}(x) \text{ then }\cos(y) = x \\ \text{We can find dy/dx by differentiating the latter implicitly as done above.} \\ \frac{d}{dx}\cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}} \]
thank you ! that makes more sense to me now!
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