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How do you differentiate #f(x)=cotx# using the quotient rule?

Answer 1

Ezekiel Alexander

# f'(x)=-csc^2x#.

#f(x)=cotx=cosx/sinx#

#:. f'(x)=d/dx(cosx/sinx)= (sinx(d/dxcosx)-cosxd/dx(sinx))/(sinx)^2#

#={sinx(-sinx)-cosx(cosx)}/sin^2x#

#=-(sin^2x+cos^2x)/sin^2x=-1/sin^2x#

#-csc^2x#.

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Answer 2

William Abdalla

To differentiate ( f(x) = \cot(x) ) using the quotient rule:

Identify ( u(x) ) and ( v(x) ) as follows: ( u(x) = 1 ) ( v(x) = \tan(x) )
Apply the quotient rule: ( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} )
Differentiate ( u(x) ) and ( v(x) ): ( u'(x) = 0 ) ( v'(x) = \sec^2(x) )
Substitute into the quotient rule formula: ( f'(x) = \frac{0 \cdot \tan(x) - 1 \cdot \sec^2(x)}{(\tan(x))^2} )
Simplify: ( f'(x) = -\frac{\sec^2(x)}{\tan^2(x)} )
Using trigonometric identities: ( f'(x) = -\csc^2(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.