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tangent f(x)=(sqrt(2))^x,\at x=6

解答

切线

f (x) = (2)^{x}, at x = 6

解答

y = 4 ln (2) x + 8 - 24 ln (2)

求解步骤

找到切点:

(6, 8)

计算

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

的斜率:

\frac{df ( x )}{d x} = \frac{1}{2} ln (2) e^{\frac{l n ( 2 ) x}{2}}

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 = 4 ln (2)

找到斜率 m=

4 ln (2)

且穿过

(6, 8)

的直线:

y = 4 ln (2) x + 8 - 24 ln (2)

y = 4 ln (2) x + 8 - 24 ln (2)

作图

流行的例子

y^{''}+14y^'+28y=6675e^{2t}cos(15t)

y^{^{''}} + 14 y^{^{'}} + 28 y = 6675 e^{2 t} cos (15 t)

laplacetransform t^2sin(t)

l a pl a ce t r an s f or m t^{2} sin (t)

derivative (1+5t)/(ln(t^2))

d er i v a t i v e \frac{1 + 5 t}{ln ( t ^{2} )}

integral of (x^2)/(sqrt(x^6-1))

\int \frac{x ^{2}}{x ^{6} - 1} d x

integral from 1 to 10 of 5x^{-2}

\int_{1}^{10} 5 x^{- 2} d x