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Question 1

Question 2

Решите задачу:

$17)\; \; \{\; x=ln\frac{1}{\sqrt{1-t^4}} \; ,\; \; y=arcsin\frac{1-t^2}{1+t^2}\; \}\\\\x'_{t}=\sqrt{1-t^4}\cdot \frac{-\frac{-4t^3}{2\sqrt{1-t^4}}}{1-t^4}=\frac{2t^3}{1-t^4}\; ,\\\\y'_{t}=\frac{1}{\sqrt{1-(\frac{1-t^2}{1+t^2})^2}}\cdot \frac{-2t(1+t^2)-2t(1-t^2)}{(1+t^2)^2}=\frac{1}{\sqrt{\frac{1+2t^2+t^4-1+2t^2-t^4}{(1+t^2)^2}}}\cdot \frac{-2t-2t^3-2t+2t^3}{(1+t^2)^2}=\\\\=\frac{1+t^2}{\sqrt{4t^2}}\cdot \frac{-4t}{(1+t^2)^2}=-\frac{2}{1+t^2}\\\\y'_{x}=\frac{y'_{t}}{x'_{t}}=-\frac{2}{1+t^2}\cdot \frac{1-t^4}{2t^3}=-\frac{(1-t^2)(1+t^2)}{(1+t^2)\cdot t^3}=-\frac{1-t^2}{t^3}$

$2)\; \; \{\; x=ln(ctgt)\; \; ,\; \; y=\frac{1}{cos^2t}\; \}\\\\x'_{t}=\frac{1}{ctgt}\cdot \frac{-1}{sin^2t}=-\frac{sint}{cost\cdot sin^2t}=-\frac{1}{sint\cdot cost}=-\frac{2}{sin2t}\\\\y'_{t}=\frac{-2cost\cdot (-sint)}{cos^4t}=\frac{sin2t}{cos^4t}\\\\y'_{x}=\frac{sin2t}{cos^4t}\cdot \frac{sin2t}{-2}=-\frac{sin^22t}{2cos^4t}$

$3)\; \; \{x=ln(t+\sqrt{t^2+1)}\; \; ,\; \; y=t\sqrt{t^2+1}\; \}\\\\x'_{t}=\frac{1}{t+\sqrt{t^2+1}}\cdot (1+\frac{2t}{2\sqrt{t^2+1}})=\frac{1}{t+\sqrt{t^2+1}}\cdot \frac{\sqrt{t^2+1}+t}{\sqrt{t^2+1}}=\frac{1}{\sqrt{t^2+1}}\\\\y'_{t}=\sqrt{t^2+1}+t\cdot \frac{2t}{2\sqrt{t^2+1}}=\frac{t^2+1+2t^2}{\sqrt{t^2+1}}=\frac{3t^2+1}{\sqrt{t^2+1}}\\\\y'_{x}=\frac{3t^2+1}{\sqrt{t^2+1}}\cdot \frac{\sqrt{t^2+1}}{1}=3t^2+1$