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受欢迎的微积分 >

tangent f(x)=e^{5x},\at x= 1/5 ln(2)

解答

切线

f (x) = e^{5 x}, at x = \frac{1}{5} ln (2)

解答

y = 10 x + 2 - 2 ln (2)

求解步骤

找到切点:

(\frac{1}{5} ln (2), 2)

计算

f (x) = e^{5 x}

的斜率:

\frac{df ( x )}{d x} = e^{5 x} \cdot 5

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 = 10

找到斜率 m=

10

且穿过

(\frac{1}{5} ln (2), 2)

的直线:

y = 10 x + 2 - 2 ln (2)

y = 10 x + 2 - 2 ln (2)

作图

流行的例子

(\partial)/(\partial x)(sqrt(96-5x^2-3y^2))

\frac{\partial}{\partial x} (96 - 5 x^{2} - 3 y^{2})

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t an g e n t f (x) = x^{3} + 1

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\int \frac{e ^{x}}{e ^{3 x} + e ^{2 x}} d x

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\frac{\partial}{\partial y} (∣ y ∣ + \frac{1}{2 a} ∣ x - y ∣^{2})

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\int \frac{1}{4 x + 1} d x