Question

найти производную обратной тригонометрической функции

Answer 1

$y' = (arccos(ctg^{2} x))' = - frac{1}{ sqrt{1 - ctg^{4} x} } times (ctg^{2} x)' = - frac{1}{ sqrt{1 - ctg^{4} x} } times2ctgx times ( - frac{1}{sin^{2}x}) = frac{2ctgx}{sin^{2}x sqrt{1 - ctg^{4} x} } = frac{2cosx}{sinx times sin^{2}x times frac{ sqrt{ sin^{4} x- cos^{4}x} }{sin^{2}x} } = frac{2cosx}{sinx sqrt{sin^{2}x - cos^{2}x} } = frac{2ctgx}{sqrt{sin^{2}x - cos^{2}x} }$

Answer 2

$y=arccos(ctg^2x); ; ,; ; ; (arccosu)'=-frac{1}{sqrt{1-u^2}}cdot u'; ; ,; ; u=ctg^2x\\y'=-frac{1}{sqrt{1-(ctg^2x)^2}}cdot (ctg^2x)'=Big [; (u^2)'=2ucdot u'; ,; ; u=ctgx; Big ]=\\=-frac{1}{sqrt{1-ctg^4x}}cdot 2, ctgxcdot (-frac{1}{sin^2x})=frac{2ctgx}{sin^2xcdot sqrt{1-frac{cos^4x}{sin^4x}}}=\\=frac{2ctgx}{sin^2xcdot sqrt{frac{sin^4x-cos^4x}{sin^4x}}}=frac{2ctgx}{sin^2xcdot frac{sqrt{(sin^2x-cos^2x)(sin^2x+cos^2x)}}{sin^2x}}=$

$=frac{2ctgx}{sqrt{sin^2x-cos^2x}}=frac{2ctgx}{sqrt{-(cos^2x-sin^2x)}}=frac{2ctgx}{sqrt{-cos2x}}$