If #f(x)= cot5 x # and #g(x) = 2x^2 1 #, how do you differentiate #f(g(x)) # using the chain rule?
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To differentiate ( f(g(x)) ) using the chain rule, follow these steps:

Compute the derivative of the outer function ( f(x) ) with respect to its inner function ( g(x) ). This is denoted as ( f'(g(x)) ).

Compute the derivative of the inner function ( g(x) ) with respect to ( x ), denoted as ( g'(x) ).

Multiply ( f'(g(x)) ) by ( g'(x) ).

Substitute ( g(x) ) back into the resulting expression.
Applying this to ( f(g(x)) = \cot^5(g(x)) ) and ( g(x) = 2x^2  1 ):

Compute ( f'(g(x)) ) by finding the derivative of ( \cot^5(x) ), which is ( 5\cot^4(x)\csc^2(x) ).

Compute ( g'(x) ) by finding the derivative of ( 2x^2  1 ), which is ( 4x ).

Multiply ( f'(g(x)) ) by ( g'(x) ) to get ( 20x\cot^4(2x^2  1)\csc^2(2x^2  1) ).
Therefore, the derivative of ( f(g(x)) ) with respect to ( x ) using the chain rule is ( 20x\cot^4(2x^2  1)\csc^2(2x^2  1) ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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