Question

Будь ласка обчисліть похідні.

Answer 1

Ответ:

$1)\ \ y=ln(tg^33x)\\\\y'=\dfrac{1}{tg^33x}\cdot 3\, tg^23x\cdot \dfrac{1}{cos^23x}\cdot 3=\dfrac{9}{tg3x\cdot cos^23x}=\dfrac{9}{sin3x\cdot cos3x}=\dfrac{18}{sin6x}\\\\\\2)\ \ y=x+\dfrac{1}{x+\sqrt{1+x^2}}\\\\\\y'=1-\dfrac{1+\frac{x}{\sqrt{1+x^2}}}{(x+\sqrt{1+x^2})^2}=1-\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}\, (x+\sqrt{1+x^2})}=1-\dfrac{1}{\sqrt{1+x^2}}$

$3)\ \ y=cos\Big(ln\sqrt{x+\dfrac{1}{x}}\ \Big)\\\\\\y'=-sin\Big(ln\sqrt{x+\dfrac{1}{x}}\ \Big)\cdot \dfrac{1}{\sqrt{x+\dfrac{1}{x}}}\cdot \dfrac{1}{2\sqrt{x+\dfrac{1}{x}}}\cdot \Big(1-\dfrac{1}{x^2}\Big)=\\\\\\=-sin\Big(ln\sqrt{x+\dfrac{1}{x}}\ \Big)\cdot \dfrac{x^2-1}{2x(x^2+1)}$

$4)\ \ cos(2x-3y)+2x-4y=0\\\\-sin(2x-3y)\cdot (2-3yy')+2-4y'=0\\\\-2sin(2x-3y)+3yy'\, sin(2x-3y)+2-4y'=0\\\\y'\cdot (3y\, sin(2x-3y)-4)=2sin(2x-3y)-2\\\\y'=\dfrac{2sin(2x-3y)-2}{2y\, sin(2x-3y)-4}$  

$5)\ \ \left\{\begin{array}{l}x=e^{3t}+t\\y=arctg(2t)\end{array}\right\ \ \ \ \ \ \ y'_{x}=\dfrac{y'_{t}}{x'_{t}}\\\\\\y'_{x}=\dfrac{\dfrac{2}{1+4t^2}}{3e^{3t}}=\dfrac{2}{3e^{3t}(1+4t^2)}$