# What is the derivative of #f(x)=(cos^-1(x))/x# ?

Using Quotient Rule, which is

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To find the derivative of ( f(x) = \frac{\cos^{-1}(x)}{x} ), we can use the quotient rule. The quotient rule states that if ( u(x) ) and ( v(x) ) are differentiable functions, then the derivative of ( \frac{u(x)}{v(x)} ) is given by:

[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]

In this case, ( u(x) = \cos^{-1}(x) ) and ( v(x) = x ). To differentiate ( \cos^{-1}(x) ), we can use the chain rule, which states that if ( f(x) ) is differentiable and ( g(x) ) is differentiable at ( f(x) ), then the derivative of ( g(f(x)) ) is ( g'(f(x)) \cdot f'(x) ).

Therefore, the derivative of ( \cos^{-1}(x) ) is ( -\frac{1}{\sqrt{1-x^2}} ) (you can verify this using the chain rule).

Now, applying the quotient rule, we have:

[ f'(x) = \frac{-\frac{1}{\sqrt{1-x^2}} \cdot x - \cos^{-1}(x) \cdot 1}{x^2} ]

Simplifying, we get:

[ f'(x) = \frac{-x}{x\sqrt{1-x^2}} - \frac{\cos^{-1}(x)}{x^2} ]

[ f'(x) = -\frac{1}{\sqrt{1-x^2}} - \frac{\cos^{-1}(x)}{x^2} ]

So, the derivative of ( f(x) = \frac{\cos^{-1}(x)}{x} ) is ( -\frac{1}{\sqrt{1-x^2}} - \frac{\cos^{-1}(x)}{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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