評估
\frac{\sin(t)\sin(2t)\sin(t\cos(t))-2\left(\sin(t)\right)^{3}\cos(t\cos(t))+2\sin(t)\sin(t\cos(t))+2\cos(t)\cos(t\cos(t))}{2\sin(t\left(-\cos(t)+1\right))}
對 t 微分
\frac{-2t\sin(t)\left(\cos(t)\sin(t\cos(t))\right)^{2}-2t\sin(t)\left(\cos(t)\cos(t\cos(t))\right)^{2}+2\cos(t)\left(\sin(t)\sin(t\cos(t))\right)^{2}+2\cos(t)\left(\sin(t)\cos(t\cos(t))\right)^{2}-4\sin(t)\left(\cos(t)\right)^{3}\left(\sin(t\cos(t))\right)^{2}-4\cos(t)\left(\sin(t)\right)^{3}\left(\cos(t\cos(t))\right)^{2}-2\left(\sin(t)\sin(t\cos(t))\right)^{2}-2\left(\sin(t)\cos(t\cos(t))\right)^{2}-2\left(\cos(t)\sin(t\cos(t))\right)^{2}-2\left(\cos(t)\cos(t\cos(t))\right)^{2}+\left(\sin(2t)\right)^{2}\sin(2t\cos(t))-2t\left(\sin(t)\right)^{3}+2\left(\cos(t)\right)^{3}}{2\left(\sin(t\left(-\cos(t)+1\right))\right)^{2}}
共享
已復制到剪貼板
示例
二次方程式
{ x } ^ { 2 } - 4 x - 5 = 0
三角學
4 \sin \theta \cos \theta = 2 \sin \theta
線性方程
y = 3x + 4
算術
699 * 533
矩陣
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
聯立方程
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
微分
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
積分
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
限制
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}