Question

Найти производную функции y=√x^2+sin^2(3x)

Answer 1

y'(x)= (x²+sin3x)'=

=(x²)'+(sin3x)'=

=2x+cos3x•(3x)'=

=2x+3cos3x

Пошаговое объяснение:

Answer 2

$y = \sqrt{ {x}^{2} + { \sin }^{2} (3x)}$

$y' = ( {( {x}^{2} + { \sin }^{2} (3x) )}^{ \frac{1}{2} } )' = \\ = \frac{1}{2} {( {x}^{2} + { \sin }^{2}(3x) ) }^{ - \frac{1}{2} } \times ( {x}^{2} + { \sin }^{2} (3x))' = \\ = \frac{1}{2 \sqrt{ {x}^{2} + { \sin }^{2}(3x) } } \times (2x + 2 \sin(3x) \times ( \sin(3x) )' \times (3x)') = \\ = \frac{2x + 2 \sin(3x) \times \cos(3x) \times 3}{2 \sqrt{ {x}^{2} + { \sin }^{2} (3x) } } = \\ = \frac{2x + 3 \sin(6x) }{2 \sqrt{ {x}^{2} + { \sin}^{2}(3x) } }$