1.次の方程式で定められる\(x\)の関数\(y\)の導関数\(\displaystyle \frac{dy}{dx}\)を求めなさい。
(1)\(x^2+y^2=1\)
両辺を\(x\)で微分すると、
\(\displaystyle 2x+\frac{dy}{dx}2y=0\)
\(\displaystyle \frac{dy}{dx}=-\frac{x}{y}\)
(2)\(x^2-y^2=1\)
両辺を\(x\)で微分すると、
\(\displaystyle 2x-\frac{dy}{dx}2y=0\)
\(\displaystyle \frac{dy}{dx}=\frac{x}{y}\)
2.\(x,y\)が媒介変数\(t\)を用いて表されるとき、\(x\)の関数\(y\)の導関数\(\displaystyle \frac{dy}{dx}\)を\(t\)を用いて求めなさい。
(1)\(x=t^2,y=t^3\)
\(\displaystyle \frac{dx}{dt}=2t\)
\(\displaystyle \frac{dy}{dt}=3t^2\)
\(\displaystyle \frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3}{2}t\)
(2)\(x=2t^2,y=2t-1\)
\(\displaystyle \frac{dx}{dt}=4t\)
\(\displaystyle \frac{dy}{dt}=2\)
\(\displaystyle \frac{dy}{dx}=\frac{2}{4t}=\frac{1}{2t}\)
(3)\(x=\cos t,y=\sin t\)
\(\displaystyle \frac{dx}{dt}=-\sin t\)
\(\displaystyle \frac{dy}{dt}=\cos t\)
\(\displaystyle \frac{dy}{dx}=-\frac{\cos t}{\sin t}\)