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How do you differentiate #f(x) = sin^2 x + 1/2 cot x-tanx#?

Answer 1

Samantha Adkison

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

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

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

#f'(x)=sin(2x)-1/(2*sin^2(x))-1/cos^2(x)#

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Answer 2

Ezekiel Alexander

To differentiate ( f(x) = \sin^2(x) + \frac{1}{2}\cot(x) - \tan(x) ), you can differentiate each term separately using the rules of differentiation.

The derivative of ( \sin^2(x) ) with respect to ( x ) is ( 2\sin(x)\cos(x) ).

The derivative of ( \frac{1}{2}\cot(x) ) with respect to ( x ) is ( -\frac{1}{2}\csc^2(x) ).

The derivative of ( -\tan(x) ) with respect to ( x ) is ( -\sec^2(x) ).

Therefore, the derivative of ( f(x) = \sin^2(x) + \frac{1}{2}\cot(x) - \tan(x) ) with respect to ( x ) is:

[ f'(x) = 2\sin(x)\cos(x) - \frac{1}{2}\csc^2(x) - \sec^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.