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How do you differentiate #f(x)=cot(5x)+csc(5x)#?

Answer 1

Charles Anctil

#f'(x) = -5csc(5x)(csc(5x)+cot(5x))#

#f(x) = cot(5x) + csc(5x)#

#f'(x) = d/dx(cot(5x)) +d/dx(csc(5x))#

Applying standard differentials and the chain rule to each term:

#f'(x )# = #-csc^2 (5x) * d/dx(5x)# #- cot(5x) * csc(5x)# * #d/dx(5x)#

#= -csc^2 (5x) * 5 - cot(5x) * csc(5x) * 5#

#=-5csc(5x)(csc(5x)+cot(5x))#

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Answer 2

Ezra Adams

To differentiate the function ( f(x) = \cot(5x) + \csc(5x) ), you would use the differentiation rules for cotangent and cosecant functions. The derivatives of cotangent and cosecant are:

[ \frac{d}{dx}(\cot(x)) = -\csc^2(x) ] [ \frac{d}{dx}(\csc(x)) = -\csc(x) \cot(x) ]

Applying these derivatives to the given function:

[ \frac{d}{dx}(f(x)) = \frac{d}{dx}(\cot(5x)) + \frac{d}{dx}(\csc(5x)) ] [ = -5\csc^2(5x) - 5\csc(5x) \cot(5x) ]

So, the derivative of ( f(x) ) is ( -5\csc^2(5x) - 5\csc(5x) \cot(5x) ).

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Answer from HIX Tutor

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