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

1 Answer

Jun 1, 2018

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

Explanation:

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)#

so

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

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