Solve \left\{ \begin{array} { l } { x = \sqrt { t ^ {

求值

{x = t^{2} + 1 y = ln (t + t^{2} + 1)

求导数 \frac{d y}{d x} ，先求 \frac{d x}{d t} 和 \frac{d y}{d t}

\frac{d}{d t} (x) = \frac{d}{d t} (t^{2} + 1) \frac{d}{d t} (y) = \frac{d}{d t} (ln (t + t^{2} + 1))

求导数

更多步骤

\frac{d x}{d t} = \frac{t}{t ^{2} + 1} \frac{d}{d t} (y) = \frac{d}{d t} (ln (t + t^{2} + 1))

求导数

更多步骤

\frac{d x}{d t} = \frac{t}{t ^{2} + 1} \frac{d y}{d t} = \frac{1}{t ^{2} + 1}

通过将 \frac{d x}{d t} = \frac{t}{t ^{2} + 1} 和 \frac{d y}{d t} = \frac{1}{t ^{2} + 1} 代入 \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} 找到所需的导数

\frac{d y}{d x} = \frac{\frac{1}{t ^{2} + 1}}{\frac{t}{t ^{2} + 1}}

解题方案

更多步骤

\frac{d y}{d x} = \frac{1}{t}

\left\{ \begin{array} { l } { x = \sqrt { t ^ { 2 } + 1 } } \\ { y = \ln ( t + \sqrt { t ^ { 2 } + 1 } ) } \end{array} \right.