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受欢迎的微积分 >

tangent y=(sqrt(5))^x

解答

切线

y = (5)^{x}

解答

y = \frac{1}{2} ln (5) e^{\frac{l n ( 5 ) a _{0}}{2}} x + 5^{\frac{a _{0}}{2}} - \frac{1}{2} ln (5) e^{\frac{l n ( 5 ) a _{0}}{2}} a_{0}

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, 5^{\frac{a _{0}}{2}})

计算

y = (5)^{x}

的斜率:

\frac{d y}{d x} = \frac{1}{2} ln (5) e^{\frac{l n ( 5 ) 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 = \frac{1}{2} ln (5) e^{\frac{l n ( 5 ) a _{0}}{2}}

找到斜率 m=

\frac{1}{2} ln (5) e^{\frac{l n ( 5 ) a _{0}}{2}}

且穿过

(a_{0}, 5^{\frac{a _{0}}{2}})

的直线:

y = \frac{1}{2} ln (5) e^{\frac{l n ( 5 ) a _{0}}{2}} x + 5^{\frac{a _{0}}{2}} - \frac{1}{2} ln (5) e^{\frac{l n ( 5 ) a _{0}}{2}} a_{0}

y = \frac{1}{2} ln (5) e^{\frac{l n ( 5 ) a _{0}}{2}} x + 5^{\frac{a _{0}}{2}} - \frac{1}{2} ln (5) e^{\frac{l n ( 5 ) a _{0}}{2}} a_{0}

作图

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