Question

производнаябыстрее пожалуйста​

Answer 1

Ответ:

$1) 2x+1\\2) 30x\\3) 12x^2 - 14x\\4)10x^4+8x\\5) 4x^3 - \frac{1}{x^2}\\6)\frac{2}{\sqrt{x} } \\7)5cos(x)\\8) -\frac{4}{sin^2(x)} \\9) cos(x) - (x+2)sin(x)\\10)\frac{x^2+12x+4}{(x+6)^2}$

Объяснения:

$1) (x^2+x)' = (x^2)' + (x)' = 2x^{2-1} + 1x^{1-1} = 2x^{1} + 1x^{0}= 2x+1\\2) (15x^2+36)' = (15x^2)' + (36)' = 2*15x+0=30x\\3) (4x^3-7x^2+2)' = (4x^3)'-(7x^2)'+(2)' = 3*4x^2-2*7x+0 = 12x^2-14x\\4)(2x^5 + 4x^2)' = 5*2x^4 + 2*4x=10x^4+8x\\\\5)(x^4 + \frac{1}{x})' = 4x^3+(x^{-1})' = 4x^3 + (-1)x^{-2} = 4x^3 - \frac{1}{x^2}\\\\6) (4\sqrt{x} +4 )'= (4\sqrt{x} )' + 0 = 4(x^{\frac{1}{2} })' = 4*\frac{1}{2} x^{\frac{1}{2} - 1 } = 2 x^{-\frac{1}{2}} = \frac{2}{\sqrt{x} }$

$7) (5sin(x)+4)' = 5(sin(x))' + 0 = 5cos(x)\\8)(4ctg(x) - 3)' = 4(ctg(x))' - 0 = -\frac{4}{sin^2(x)} \\9) ((x+2)cos(x))' = (x+2)'cos(x) + (x+2)(cos(x))' = cos(x) - (x+2)sin(x)\\\\10) \frac{x^2-4}{x+6} = \frac{(x^2-4)'(x+6)-(x^2-4)(x+6)'}{(x+6)^2} =\frac{2x(x+6)-(x^2-4)}{(x+6)^2} = \frac{2x^2+12x-x^2+4}{(x+6)^2} =\frac{x^2+12x+4}{(x+6)^2}$