Question

Мне нужно, а не знаю как его решать, , любой пример

Answer 1

$\left \{ {{x=\sqrt{1-4^{t}}} \atop {y=arcsin2^{t}}} \right.\\\\x'_{t}=\frac{1}{2\sqrt{1-4^{t}}}\cdot (1-4^{t})'=\frac{1}{2\sqrt{1-4^{t}}}\cdot (-4^{t}\cdot ln4)=-\frac{4^{t}\cdot ln2}{\sqrt{1-4^{t}}}\\\\y'_{t}=\frac{1}{\sqrt{1-(2^{t})^2}}\cdot (2^{t})'=\frac{1}{\sqrt{1-4^{t}}}\cdot 2^{t}\cdot ln2=\frac{2^{t}\cdot ln2}{\sqrt{1-4^{t}}}\\\\\frac{dy}{dx}=y'_{x}=\frac{y'_{t}}{x'_{t}}=-\frac{2^{t}\cdot ln2}{4^{t}\cdot ln2}=-\frac{1}{2^{t}}$

$y''_{tt}=(y'_{t})'=\Big (\frac{2^{t}\cdot ln2}{\sqrt{1-4^{t}}}\Big )'=\frac{2^{t}\cdot ln^22\cdot \sqrt{1-4^{t}}-2^{t}\cdot ln2\cdot \frac{1}{2\sqrt{1-4^{t}}}\cdot 4^{t}\cdot ln4}{1-4^{t}}=\\\\=\frac{2^{t}\cdot ln^22\cdot (1-4^{t})-2^{t}\cdot ln2\cdot 4^{t}\cdot ln2}{\sqrt{1-4^{t}}\cdot (1-4^{t})}=\frac{2^{t}\cdot ln^22\cdot (1-2\cdot 4^{t})}{\sqrt{(1-4^{t})^3}}\\\\\frac{d^2y}{dx^2}=y''_{xx}=\frac{y''_{tt}}{x'_{t}}=\frac{2^{t}\cdot ln^22\cdot (1-2\cdot 4^{t})}{-4^{t}\cdot ln2\cdot (1-4^{t})}=-\frac{ln2\cd(1-2\cdot 4^{t})}{2^{t}\cdot (1-4^{t})}$