Question

продифференцировать функции
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Answer 1

Ответ:

$y '= \frac{U'V - V'U}{ {V}^{2} } = \\ = \frac{2th(x + 3) \times \frac{1}{ {ch}^{2} (x + 3) } \times arcctg \sqrt{x} - ( - \frac{1}{1 + x} ) \times \frac{1}{2 \sqrt{x} } \times {th}^{2}(x + 3) }{ {arcctg}^{2} \sqrt{x} } = \\ = \frac{ \frac{2th(x + 3) \times arcctg \sqrt{x} }{ {ch}^{2}(x + 3) } + \frac{ {th}^{2} (x + 3)}{2 \sqrt{x}(1 + x) } }{ {arcctg}^{2} \sqrt{x} } = \\ = \frac{2th(x + 3) }{arcctg \sqrt{x} \times {ch}^{2} (x + 3)} + \frac{ {th}^{2} (x + 3)}{2 \sqrt{x} (1 + x) {arcctg}^{2}(x + 1) }$