Answer:
[tex]\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1-x^2}}[/tex]
[tex]\dfrac{dy}{dx} = -\csc(y)[/tex]
Step-by-step explanation:
First, we can isolate y:
[tex]x=\cos(y)[/tex]
↓ taking the inverse cosine of both sides
[tex]\cos^{-1}(x) = y[/tex]
Next, we can use the common derivative to differentiate both sides:
[tex](\cos^{-1}(x))' = \dfrac{-1}{\sqrt{1-x^2}}[/tex]
↓↓↓
[tex]\dfrac{d}{dx}(y)=\dfrac{d}{dx}(\cos^{-1}(x))[/tex]
[tex]\boxed{\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1-x^2}}}[/tex]
Note that we can substitute back in the definition for x in terms of y to get a derivative in terms of y:
[tex]\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1-(\cos(y))^2}}[/tex]
↓ applying the Pythagorean identity: [tex](\sin(\theta))^2=1-(\cos(\theta))^2[/tex]
[tex]\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{(\sin(y))^2}}[/tex]
[tex]\dfrac{dy}{dx} = \dfrac{-1}{\sin(y)}[/tex]
↓ applying the definition of [tex]\csc(\theta) = \dfrac{1}{\sin(\theta)}[/tex]
[tex]\boxed{\dfrac{dy}{dx} = -\csc(y)}[/tex]