tangent f(x)=x^2+4,(-3,13)

解答

切线

f (x) = x^{2} + 4, (- 3, 13)

y = - 6 x - 5

求解步骤

计算

f (x) = x^{2} + 4

的斜率:

\frac{df ( x )}{d x} = 2 x

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 = - 6

找到斜率 m=

- 6

且穿过

(- 3, 13)

的直线:

y = - 6 x - 5

y = - 6 x - 5

\int \frac{cos ^{3} ( 3 t )}{sin ^{5} ( 3 t )} d t

y^{^{'}} + \frac{2 y}{t - 2} = \frac{sin ( t )}{t - 2}

\frac{\partial}{\partial x} (3 x y + 4 x z + 5 yz - 3)

\frac{d}{d x} (3 cot (x))

\int 2 - 2 cos (t) d t