How do you differentiate #f(x) = sqrt(arctan(e^(x1)) # using the chain rule?
This is a pretty big composition  the square root, inverse tangent, and exponential will all have to be differentiated.
Differentiate the square root:
So, we have
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To differentiate ( f(x) = \sqrt{\arctan(e^{x1})} ) using the chain rule, you would follow these steps:

Identify the outer function ( u ) and the inner function ( v ).
 Let ( u = \sqrt{v} ) where ( v = \arctan(e^{x1}) ).

Find the derivatives of ( u ) and ( v ) with respect to ( x ).
 ( \frac{du}{dv} = \frac{1}{2\sqrt{v}} )
 ( \frac{dv}{dx} = \frac{1}{1 + (e^{x1})^2} \cdot e^{x1} )

Apply the chain rule: ( \frac{df}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} ).
 ( \frac{df}{dx} = \frac{1}{2\sqrt{v}} \cdot \frac{1}{1 + (e^{x1})^2} \cdot e^{x1} )

Substitute the expressions for ( u ) and ( v ) back into the equation.
 ( \frac{df}{dx} = \frac{1}{2\sqrt{\arctan(e^{x1})}} \cdot \frac{1}{1 + (e^{x1})^2} \cdot e^{x1} )
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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