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tangent f(x)=sqrt(3x+7),\at x=6

解答

切线

f (x) = 3 x + 7, at x = 6

解答

y = \frac{3}{10} x + \frac{16}{5}

求解步骤

找到切点:

(6, 5)

计算

f (x) = 3 x + 7

的斜率:

\frac{df ( x )}{d x} = \frac{3}{2 3 x + 7}

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{3}{10}

找到斜率 m=

\frac{3}{10}

且穿过

(6, 5)

的直线:

y = \frac{3}{10} x + \frac{16}{5}

y = \frac{3}{10} x + \frac{16}{5}

作图

流行的例子

derivative (1-3x)/(6+x)

d er i v a t i v e \frac{1 - 3 x}{6 + x}

derivative f(x)= 1/((x+1)^2)

d er i v a t i v e f (x) = \frac{1}{( x + 1 ) ^{2}}

sum from n=1 to infinity of n/(2^n)

n = 1 \sum \infty \frac{n}{2 ^{n}}

slopeintercept (6,-1),(-3,-10)

s l o p e in t erce pt (6, - 1), (- 3, - 10)

d/(dy)((y^3)/(30)+5/(2y))

\frac{d}{d y} (\frac{y ^{3}}{30} + \frac{5}{2 y})