Question

Срочно!!!!!!!!! Помогите ​

Answer 1

$1)\; \; y=x^3+7x^9-17\; \; ,\; \; y'=3x^2+63x^8\\\\y=2\sqrt{x}-\frac{4}{x}-\frac{x}{7}\; \; ,\; \; y'=\frac{1}{\sqrt{x}}+\frac{4}{x^2}-\frac{1}{7}\\\\y=4cosx-7tgx\; \; ,\; \; y'=-4sinx-\frac{7}{cos^2x}\\\\y=4e^{x}-8log_5x\; \; ,\; \; y'=4e^{x}-\frac{8}{x\, ln5}$

$5)y=(x^8+3)(2x^4-27)\; \; ,\\\\y'=8x^7(2x^4-27)+8x^3(x^8+3)=24x^{11}-216x^7+24x^3\\\\6)\; \; y=(12x-x^2)\cdot \sqrt{x}\; \; ,\; \; y=12x^{\frac{3}{2}}-x^{\frac{5}{2}}\\\\y'=12\cdot \frac{3}{2}x^{\frac{1}{2}}-\frac{5}{2}x^{\frac{3}{2}}=18\sqrt{x}-2,5\sqrt{x^3}\\\\7)\; \; y=-6x\, cosx\\\\y'=-6\, cosx-6x\cdot (-sinx)=-6\, cosx+6x\, sinx\\\\8)\; \; y=\frac{x^3}{5x^2+6}\\\\y'=\frac{3x^2(5x^2+6)-x^3\cdot 10x}{(5x^2+6)^2}=\frac{5x^4+18x^2}{(5x^2+6)^2}=\frac{x^2\cdot (5x^2+18)}{(5x^2+6)^2}$

$9)\; \; y=\frac{5sinx}{7x}\\\\y'=\frac{5\, cosx\cdot 7x-5sinx\cdot 7}{49x^2}=\frac{35\cdot (x\cdot cosx-sinx)}{49x^2}=\frac{5\cdot (x\cdot cosx-sinx)}{7x^2}\\\\10)\; \; y=\frac{3^{x}}{2x^5}\\\\y'=\frac{3^{x}\cdot ln3\cdot 2x^5-3^{x}\cdot 10x^4}{4x^{10}}=\frac{2x^4\cdot 3^{x}\cdot (x\cdot ln3-5)}{4x^{10}}=\frac{x^4\cdot 3^{x}\cdot (x\cdot ln3-5)}{2x^{10}}$