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

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

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