Question

найти производную а) и г)​

Answer 1

Ответ:

Найти производные функций .

$\bf 1)\ \ y=\dfrac{3+6x}{\sqrt{3-4x+5x^2}}$  

Формулы :   $\bf \Big(\dfrac{u}{v}\Big)'=\dfrac{u'v-uv'}{v^2}\ \ ,\ \ (\sqrt{u} )'=\dfrac{1}{2\sqrt{u}}\cdot u'$   .

$\bf y'=\dfrac{6\cdot \sqrt{3-4x+5x^2}-(3+6x)\cdot \dfrac{10x-4}{2\sqrt{3-4x+5x^2}}}{3-4x+5x^2}=\\\\\\=\dfrac{6\cdot (3-4x+5x^2)-(3+6x)\cdot (10x-4)}{\sqrt{(3-4x+5x^2)^3}}=\dfrac{30\, (1-x-x^2)}{\sqrt{3-4x+5x^2}}$  

$\bf 2)\ \ \dfrac{y}{x}=arctg\dfrac{x}{y}\ \ \ \Rightarrow \ \ \ \ y=x\cdot arctg\dfrac{x}{y}$  

$\bf y'=\Big(x\cdot arctg\dfrac{x}{y}\Big)'\\\\y'=arctg\dfrac{x}{y}+x\cdot \dfrac{1}{1+\dfrac{x^2}{y^2}}\cdot \dfrac{y-x\cdot y'}{y^2}\\\\y'=arcstg\dfrac{x}{y}+\dfrac{x\cdot y^2}{x^2+y^2}\cdot \Big(\dfrac{1}{y}-\dfrac{x}{y^2}\cdot y'\Big)\\\\\\y'+\dfrac{x\cdot y^2}{x^2+y^2}\cdot \dfrac{x}{y^2}\cdot y'=arctg\dfrac{x}{y}+\dfrac{x\cdot y}{x^2+y^2}\\\\\\y'\cdot \Big(1+\dfrac{x^2}{x^2+y^2}\Big)=arctg\dfrac{x}{y}+\dfrac{x\cdot y}{x^2+y^2}$  

         

$\bf y'=\dfrac{arctg\dfrac{x}{y}+\dfrac{x\cdot y}{x^2+y^2}}{1+\dfrac{x^2}{x^2+y^2}}=y'=\dfrac{(x^2+y^2)\, arctg\dfrac{x}{y}+x\cdot y}{2x^2+y^2}$