Question

Найти производную функции
а)y=arctg√3-x/x-2
б)y=(cos x)^x^2
д) x-y+e^x arctg x=0
буду благодарна))

Answer 1

Ответ:

а)

$y = arctg( \sqrt{ \frac{3 - x}{x - 2} } )$

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

б)

$y= {( \cos(x)) }^{ {x}^{2} }$

фориула:

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

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

$y' = { ( \cos(x) )}^{ {x}^{2} } \times (2x ln( \cos(x) ) - {x}^{2} \tg(x))$

в)

$x - y + {e}^{x} arctgx = 0$

$1 - y' + {e}^{x} arctgx + \frac{ {e}^{x} }{1 + {x}^{2} } = 0 \\ y' = 1 + {e}^{x} arctgx + \frac{ {e}^{x} }{1 + {x}^{2} }$