tangent y=x^2+x,(0,0)

解答

切线

y = x^{2} + x, (0, 0)

y = x

求解步骤

计算

y = x^{2} + x

的斜率:

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 1

找到斜率 m=

1

且穿过

(0, 0)

的直线:

y = x

y = x

\int_{0}^{1} \frac{22}{4 y - 1} d y

\frac{d}{d x} (\frac{x ^{2} - 1}{x ^{2} - 4})

\int \frac{cos ( y )}{sin ^{2} ( y )} d y

\frac{\partial}{\partial x} (5 e^{y})

n = 1 \sum \infty \frac{1}{( n ^{2} + 8 n )}