Question

Помогите пожалуйста, голова кипит...

Answer 1

$1)\; \; y= \frac{7}{\sqrt[3]{x^2+3x-4}}-\sqrt[5]{2+sinx}\\\\y'=-\frac{7}{3}\cdot (x^2+3x-4)^{-\frac{4}{3}}\cdot (2x+3)- \frac{1}{5}\cdot (2+sinx)^{-\frac{4}{5}}\cdot cosx\\\\2)\; \; y=ln(ctg\, 7x^2)\\\\y'=\frac{1}{ctg\, 7x^2}\cdot \frac{-1}{sin^2\, 7x^2}\cdot 14x\\\\3)\; \; y=(e^{arcsin9x^2})^5\\\\y'=5(e^{arcsin9x^2})^4\cdot e^{arcsin9x^2}\cdot \frac{1}{\sqrt{1-81x^4}}\cdot 18x\\\\4)\; \; y=(x+3)^{cos4x}\\\\ln\, y=ln(x+3)^{cos4x}\\\\ln\, y=cos4x\cdot ln(x+3)\\\\\frac{y'}{y}=-4\, sin4x\cdot ln(x+3)+cos4x\cdot \frac{1}{x+3}$

$y'=y\cdot \Big (-4\, sin4x\cdot ln(x+3)+cos4x\cdot \frac{1}{x+3}\Big )\\\\y'=(x+3)^{cos4x}\cdot \Big (-4\, sin4x\cdot ln(x+3)+\frac{cos4x}{x+3}\Big )\\\\5)\; \; sin(x^2+y^2)=e^{1/y}\\\\cos(x^2+y^2)\cdot (2x+2y\, y')=e^{1/y}\cdot (-\frac{1}{y^2})\cdot y'\\\\2x\cdot cos(x^2+y^2)=-2y\, y'\cdot cos(x^2+y^2)- \frac{e^{1/y}}{y^2}\cdot y'\\\\-y'\cdot \Big (2y\cdot cos(x^2+y^2)+ \frac{e^{1/y}}{y^2}\Big )=2x\cdot cos(x^2+y^2)\\\\y'=-\frac{2x\cdot cos(x^2+y^2)}{2y\cdot cos(x^2+y^2)+\frac{e^{1/y}}{y^2}}\\\\y'=-\frac{2xy^2\cdot cos(x^2+y^2)}{2y^3\cdot cos(x^2+y^2)+e^{1/y}}$