# How do you find the derivative #y=sqrt(tan^-1 x)#?

#1. Firstly find the derivative of arctanx...

#2. Lastly, use implicit differentiation to find the derivative of y=(arctanx)^(1/2)

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To find the derivative of ( y = \sqrt{\tan^{-1}(x)} ), you can use the chain rule.

Let ( u = \tan^{-1}(x) ), then ( y = \sqrt{u} ).

First, find the derivative of ( u ) with respect to ( x ): [ \frac{du}{dx} = \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} ]

Now, find the derivative of ( y ) with respect to ( u ): [ \frac{dy}{du} = \frac{1}{2\sqrt{u}} ]

Apply the chain rule to find the derivative of ( y ) with respect to ( x ): [ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{1}{1+x^2} ]

Substitute ( u = \tan^{-1}(x) ) back in: [ \frac{dy}{dx} = \frac{1}{2\sqrt{\tan^{-1}(x)}} \cdot \frac{1}{1+x^2} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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