найти производную y=arctg lnx/ln arctg x y=(x^3-1)^cos√x. Created by turmanfluck. matematika-ru

ln arctg x y=(x^3-1)^cos√x...

turmanfluck @turmanfluck

July 2022 1 9 Report

найти производную
y=arctg lnx/ln arctg x
y=(x^3-1)^cos√x

Answers & Comments

Miroslava227

$y = \frac{arctg (ln(x)) }{ ln(arctgx) } \\$

$y '= \frac{(arctg ln(x)) ' \times ln(arctgx) - ( ln(arctgx)) '\times arctg( ln(x)) }{ {ln}^{2} arctgx} = \\ = \frac{ \frac{1}{1 + {ln}^{2}(x) } \times \frac{1}{x} \times ln(arctgx) - \frac{1}{arctgx} \times \frac{1}{1 + {x}^{2} } \times arctg( ln(x)) }{ {ln}^{2} arctgx} = \\ = \frac{1}{ {ln}^{2} arctgx} ( \frac{ ln(arctgx) }{x(1 + {ln}^{2}(x)) } - \frac{arctg( ln(x) )}{(1 + {x}^{2} )arctgx} )$

$y = {( {x}^{3} - 1) }^{ \cos(\sqrt{x}) }$

$y '= ( ln(y)) ' \times y$

$( ln(y)) ' = ( ln( {( {x}^{3} - 1) }^{ \cos(\sqrt{x}) } ) '= ( \cos(\sqrt{x}) \times ln( {x}^{3} - 1))' = \\ = - \sin(\sqrt{x}) \times \frac{1}{2\sqrt{x}} \times ln( {x}^{3} - 1 ) + \frac{1}{ {x}^{3} - 1} \times 3 {x}^{2} \times \cos(\sqrt{x}) = \\ = - \frac{ln( {x}^{3} - 1) \sin(\sqrt{x})}{2\sqrt{x}} + \frac{3 {x}^{2} \cos(\sqrt{x}) }{ {x}^{3} - 1 }$

$y' = {( {x}^{3} - 1) }^{ \cos(\sqrt{x}) } \times ( \frac{3 {x}^{2} \cos(\sqrt{x}) }{ {x}^{3} - 1 } - \frac{ln( {x}^{3} - 1 ) \sin(\sqrt{x}))}{2\sqrt{x}} )\\$

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