Question

помогите решить пожалуйста производные! очень срочно!

Answer 1

Ответ:

5

$y = \sqrt[5]{ {x}^{2} + \sqrt[4]{2x + 1} } = {( {x}^{2} + \sqrt[4]{2x + 1} )}^{ \frac{1}{5} }$

$y '= \frac{1}{5} {( {x}^{2} + \sqrt[4]{2x + 1}) }^{ - \frac{4}{5} } \times ( {x}^{2} + {(2x + 1)}^{ \frac{1}{4} } ) '= \\ = \frac{1}{5 \sqrt[5]{ {( {x}^{2} + \sqrt[4]{2x + 1} )}^{4} } } \times (2x + \frac{1}{4} {(2x + 1)}^{ - \frac{3}{4} } \times 2) = \\ = \frac{1}{5 \sqrt[5]{ {( {x}^{2} + \sqrt[4]{2x + 1}) }^{4} } } \times (2x + \frac{1}{2 \sqrt[4]{ {(2x + 1)}^{3} } } )$

4

$y = {2}^{ {x}^{2} } tgx$

$y '= ( {2}^{ {x}^{2} } )' \times ( {x}^{2} )' \times tgx + (tgx) '\times {2}^{ {x}^{2} } = \\ = ln(2) \times {2}^{ {x}^{2} } \times 2x \times tgx + \frac{1}{ \cos {}^{2} (x) } \times {2}^{ {x}^{2} } = \\ = {2}^{ {x}^{2} } (2xtgx \times ln(2) + \frac{1}{ \cos {}^{2} (x) } )$