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How do you differentiate #f(x)=2^cotx#?

Answer 1

Luke Alvarez

#f'(x)=-2^cotxln2csc^2x#

use logarithmic differentiation

let#" "y=2^cotx#

#=>lny=cotxln2#

differentiate wrt #x#

#1/y(dy)/(dx)=- csc^2xln2#

#(dy)/(dx)=-yln2csc^2x#

substitute back for #y#

#f'(x)=-2^cotxln2csc^2x#

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Answer 2

Samantha Adkison

To differentiate ( f(x) = 2^{\cot(x)} ), you can use the chain rule for differentiation. The chain rule states that if ( g(x) ) is differentiable at ( x ) and ( f(x) ) is differentiable at ( g(x) ), then the composite function ( f(g(x)) ) is differentiable at ( x ) and its derivative is ( f'(g(x)) \cdot g'(x) ). Applying the chain rule to ( f(x) = 2^{\cot(x)} ):

( f'(x) = \ln(2) \cdot 2^{\cot(x)} \cdot (-\csc^2(x)) )

So, the derivative of ( f(x) = 2^{\cot(x)} ) is ( f'(x) = -\ln(2) \cdot \csc^2(x) \cdot 2^{\cot(x)} ).

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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.