Quod Erat Demonstrandum

2013/06/16

f4 m2 revision

Filed under: NSS — johnmayhk @ 6:39 下午
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(免插聲明:純為中四同學溫習用,高手見諒。)

同學應該頗掌握乘冪法則(Power rule):

\frac{d}{dx}(x^n)=nx^{n-1}

留意,x^n 當中的基(base)是變量 x,而冪(power)是常數 n

一旦遇上冪或/和基是變量時,比如

\frac{d}{dx}(n^x)

\frac{d}{dx}((\ln x)^x)

就不能以乘冪法則處理。

記著一招:「take log 後才 D」

e.g. 1

y=n^x
\ln y=\ln n^x
\ln y=x\ln n
\frac{d}{dx}(\ln y)=\frac{d}{dx}(x\ln n)
\frac{1}{y}\frac{dy}{dx}=\ln n
\frac{dy}{dx}=y\ln n
\therefore \frac{d}{dx}(n^x)=n^x\ln n

e.g. 2

y=(\ln x)^x
\ln y=\ln(\ln x)^x
\ln y=x\ln(\ln x)
\frac{d}{dx}(\ln y)=\frac{d}{dx}(x\ln(\ln x))
\frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}(\ln(\ln x))+\ln(\ln x)\frac{dx}{dx} (product rule)
\frac{1}{y}\frac{dy}{dx}=\frac{x}{\ln x}\frac{d}{dx}(\ln x)+\ln(\ln x) (chain rule)
\frac{1}{y}\frac{dy}{dx}=\frac{x}{\ln x}\frac{1}{x}+\ln(\ln x)
\frac{dy}{dx}=y\frac{1}{\ln x}+y\ln(\ln x)
\frac{dy}{dx}=(\ln x)^x\frac{1}{\ln x}+(\ln x)^x\ln(\ln x)
\therefore \frac{d}{dx}(\ln x)^x=(\ln x)^{x-1}+(\ln x)^x\ln(\ln x)

不過 M1 將不重視 logarithmic differentiation 這招,或用下法:

\frac{d}{dx}(n^x)
=\frac{d}{dx}e^{\ln n^x}
=\frac{d}{dx}e^{x\ln n}
=e^{x\ln n}\frac{d}{dx}(x\ln n)
=e^{\ln n^x}\ln n
=n^x\ln n

差不多吧。

習題:試以上法處理 e.g. 2。

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