tangent y=7x-x^2,(1,6)

解答

切线

y = 7 x - x^{2}, (1, 6)

y = 5 x + 1

求解步骤

计算

y = 7 x - x^{2}

的斜率:

\frac{d y}{d x} = 7 - 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 = 5

找到斜率 m=

5

且穿过

(1, 6)

的直线:

y = 5 x + 1

y = 5 x + 1

\int_{1}^{2} x^{4} + 2 x^{2} d x

\int_{0}^{1} 5 e^{x} - 5 x e^{x^{2}} d x

\int \frac{16}{( x ^{2} + 4 ) ^{2}} d x

t an g e n t y = - 3 x - 9, (- 9, 18)

d er i v a t i v e g (x) = 2 x^{0.8} + 4 x^{- 0.8}