# How do you differentiate # f(x) =arccos(2x + 1) #?

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To differentiate ( f(x) = \arccos(2x + 1) ), you can use the chain rule, which states that if ( u ) is a differentiable function of ( x ), and ( y ) is a differentiable function of ( u ), then ( y ) is a differentiable function of ( x ) and its derivative is given by ( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ).

First, identify ( u ) as the inner function, which is ( u = 2x + 1 ). Then, differentiate ( u ) with respect to ( x ), yielding ( \frac{du}{dx} = 2 ).

Next, differentiate ( y = \arccos(u) ) with respect to ( u ), resulting in ( \frac{dy}{du} = -\frac{1}{\sqrt{1-u^2}} ).

Apply the chain rule formula to find the derivative of ( f(x) ): [ \frac{df}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{\sqrt{1-(2x + 1)^2}} \cdot 2 ]

Finally, simplify the expression to get the derivative: [ \frac{df}{dx} = -\frac{2}{\sqrt{1-(2x + 1)^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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