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180100
@180100
July 2022
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y=tgx+ctgx найти производную функции и упростить
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lara164
Verified answer
Y'(X) = (tgX+ctgX)' = (tgX)'+(ctgX)' = 1/cos²X - 1/sin²X =
(sin²X-cos²X)/(cos²X*sin²X)=-4(cos²X-sin²X)/4(cosXsinX)²=
-4cos2x/(sin2Xsin2X)= -4ctg2X/sin2X
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Answers & Comments
Verified answer
Y'(X) = (tgX+ctgX)' = (tgX)'+(ctgX)' = 1/cos²X - 1/sin²X =(sin²X-cos²X)/(cos²X*sin²X)=-4(cos²X-sin²X)/4(cosXsinX)²=
-4cos2x/(sin2Xsin2X)= -4ctg2X/sin2X