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How do you find the derivative of the function: #arccos ([2x + 1]/2)#?

Answer 1

Christopher Agresta

# d/dx arccos ([2x + 1]/2) = -1/sqrt(1-(x+1/2)^2)#

We want the derivative of:

# arccos ([2x + 1]/2) = arccos (x+1/2)#

We can use the known result:

#d/dx arccos x = -1/sqrt(1-x^2)#

Along with the chain rule to get:

# d/dx arccos ([2x + 1]/2) = -1/sqrt(1-(x+1/2)^2) * d/dx (x+1/2)# # " " = -1/sqrt(1-(x+1/2)^2)#

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Answer 2

Axel Alvarez

To find the derivative of the function ( \arccos \left( \frac{2x + 1}{2} \right) ), you can use the chain rule.

Let ( u = \frac{2x + 1}{2} ). Then, ( \arccos(u) ) becomes the outer function, and ( u ) is the inner function.

The derivative of ( \arccos(u) ) with respect to ( u ) is ( -\frac{1}{\sqrt{1 - u^2}} ).

Now, using the chain rule, multiply this by the derivative of the inner function ( u ) with respect to ( x ), which is ( \frac{d}{dx}(2x + 1)/2 = \frac{1}{2} ).

Thus, the derivative of the given function is ( -\frac{1}{2\sqrt{1 - \left(\frac{2x + 1}{2}\right)^2}} ).

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Answer from HIX Tutor

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.