How do you evaluate the derivative of the function #f (x) = tan x cot x# at the indicated point (1,1)?

Answer 1
#f(x)=tanx cot x = sinx/cosx cosx/sinx = 1#
#f'(x)=0# for all #x# in the domain of #f#.
Therefore: #f'(1) = 0#
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Answer 2
I would use the Product Rule to evaluate the derivative knowing that: derivative of #tan(x)# is #1/cos^2(x)# Derivative of #cot(x)# is #-1/sin^2(x)#
(you can find these remembering that #tan(x)=sin(x)/cos(x)# and #cot(x)=cos(x)/sin(x)# and deriving using the Quotient Rule)
So you get: #f'(x)=1/cos^2(x)cot(x)+tan(x)[-1/sin^2(x)]=0#
Evaluated at #x=1# gives you: #f'(1)=0#
As an exercise on derivation is good, but you can see from the start that your function can be rearranged as: #f(x)=sin(x)/cos(x)*cos(x)/sin(x)=1# ! Which is a constant with derivative always equal to zero. graph{tan(x)cot(x) [-10, 10, -5, 5]}
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Answer 3

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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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.

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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