tangent y=sqrt(x),(16,4)

解答

切线

y = x, (16, 4)

y = \frac{1}{8} x + 2

求解步骤

计算

y = x

的斜率:

\frac{d y}{d x} = \frac{1}{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 = \frac{1}{8}

找到斜率 m=

\frac{1}{8}

且穿过

(16, 4)

的直线:

y = \frac{1}{8} x + 2

y = \frac{1}{8} x + 2

y^{^{'}} + \frac{y}{x} = \frac{7}{x ^{2}} + 3

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