Ответ:

1)   [tex]\displaystyle \bf y'=9[/tex]

2)   [tex]\displaystyle \bf y'=15x^2-4x+3[/tex]

3)    [tex]\displaystyle \bf y'=30x+7[/tex]

4)   [tex]\displaystyle \bf y'=-\frac{23}{(3x-4)^2}[/tex]

Пошаговое объяснение:

Найти производную:

1)   [tex]\displaystyle \bf y=9x-3[/tex]

[tex]\boxed {\displaystyle \bf x'=1;\;\;\;\;\;C'=0}[/tex]

[tex]\displaystyle \bf y'=9[/tex]

2)   [tex]\displaystyle \bf y=5x^3-2x^2+3x-8[/tex]

[tex]\boxed {\displaystyle \bf (x^n)'=nx^{n-1}}[/tex]

[tex]\displaystyle \bf y'=5\cdot3x^{3-1}-2\cdot2x^{2-1}+3=15x^2-4x+3[/tex]

3)   [tex]\displaystyle \bf y=(3x-1)(5x+4)[/tex]

[tex]\boxed {\displaystyle \bf (uv)'=u'b+uv'}[/tex]

[tex]\displaystyle \bf y'=(3x-1)'(5x+4)+(3x-1)(5x+4)'=3\cdot(5x+4)+(3x-1)\cdot5=\\\\15x+12+15x-5=30x+7[/tex]

4)   [tex]\displaystyle \bf y=\frac{2x+5}{3x-4}[/tex]

[tex]\boxed {\displaystyle \bf \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}}[/tex]

[tex]\displaystyle \bf y'=\frac{(2x+5)'(3x-4)-(2x+5)(3x-4)'}{(3x-4)^2} =\\\\=\frac{2\cdot(3x-4)-(2x+5)\cdot3}{(3x-4)^2} =\frac{6x-8-6x-15}{(3x-4)^2} =-\frac{23}{(3x-4)^2}[/tex]