How do you evaluate the derivative of the function #f (x) = tan x cot x# at the indicated point (1,1)?
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To evaluate the derivative of the function ( f(x) = \tan(x) \cot(x) ) at the point ( (1,1) ), we first find the derivative of the function using the product rule and then substitute ( x = 1 ) into the derivative expression.
Using the product rule, let ( u(x) = \tan(x) ) and ( v(x) = \cot(x) ). The derivative of ( u(x) ) with respect to ( x ) is ( u'(x) = \sec^2(x) ) and the derivative of ( v(x) ) with respect to ( x ) is ( v'(x) = -\csc^2(x) ).
Now applying the product rule: [ f'(x) = u'(x)v(x) + u(x)v'(x) ]
[ = (\sec^2(x))(\cot(x)) + (\tan(x))(-\csc^2(x)) ]
[ = \sec^2(x) \cot(x) - \tan(x) \csc^2(x) ]
Now, substituting ( x = 1 ): [ f'(1) = \sec^2(1) \cot(1) - \tan(1) \csc^2(1) ]
[ ≈ (1.745^2)(0.642) - (1.557)(1.321^2) ]
[ ≈ 1.122 ]
Therefore, the value of the derivative of the function ( f(x) = \tan(x) \cot(x) ) at the point ( (1,1) ) is approximately ( 1.122 ).
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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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