I'm kind of stuck on this question.

Use implicit differentiation to find the derivative of $ y =\arccos \left( {\sqrt x}\right) $ as a function of x and say where this derivative is defined.

Don't really get the concept of implicit differentiation.


$$y = \arccos \sqrt x$$

$$x = \cos^2 y$$

$$\frac{dx}{dx} = \frac{d}{dx}(\cos^2 y)$$

$$1 = \frac{d}{dy}(\cos^2 y) \cdot \frac{dy}{dx}$$

$$1 = (2)(\cos y)(-\sin y)\cdot \frac{dy}{dx}$$

$$\frac {dy}{dx} = -\frac{1}{2\cos y \sin y} = \pm\frac{1}{2\sqrt{x(1-x)}}$$

Now think about where that is defined (when is the denominator nonzero?).


Hint: $$ y=\arccos(\sqrt{x}) \quad \iff \quad \cos y= \sqrt{x} $$ now you can use implicit differentiation to find $\frac{dy}{dx}$: $$ -\sin (y) \frac{dy}{dx}=\frac{1}{2\sqrt{x}} \Rightarrow \frac{dy}{dx}=.... $$


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