How do you find the derivative of the function #f(x)=arcsin(4x)+arccos(4x)#?
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To find the derivative of ( f(x) = \arcsin(4x) + \arccos(4x) ), you can use the chain rule. The derivatives of ( \arcsin(u) ) and ( \arccos(u) ) are ( \frac{1}{\sqrt{1-u^2}} ) and ( -\frac{1}{\sqrt{1-u^2}} ), respectively. Applying the chain rule, the derivative of ( f(x) ) with respect to ( x ) is:
[ f'(x) = \frac{1}{\sqrt{1-(4x)^2}} \cdot \frac{d}{dx}(4x) + \left(-\frac{1}{\sqrt{1-(4x)^2}}\right) \cdot \frac{d}{dx}(4x) ]
[ = \frac{1}{\sqrt{1-16x^2}} \cdot 4 - \frac{1}{\sqrt{1-16x^2}} \cdot 4 ]
[ = \frac{4}{\sqrt{1-16x^2}} - \frac{4}{\sqrt{1-16x^2}} ]
[ = 0 ]
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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.
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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