Quod Erat Demonstrandum

2010/08/01

答中四同學包

Filed under: NSS — johnmayhk @ 9:22 上午

包,參考下:

1.

Find \frac{dy}{dx} of the following from first principles.

(a) y = \tan x
(b) y = xe^x

似乎有點「走火入麼」了。算,照做。

(a)

\frac{dy}{dx}
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{\tan (x + \Delta x) - \tan x}{\Delta x}
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(\frac{\sin (x + \Delta x)}{\cos (x + \Delta x)} - \frac{\sin x}{\cos x})
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x} \frac{\sin (x + \Delta x)\cos x - \cos (x + \Delta x)\sin x}{\cos (x + \Delta x) \cos x}
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x} \frac{\sin (x + \Delta x - x)}{\cos (x + \Delta x) \cos x}
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{\sin \Delta x}{\Delta x} \frac{1}{\cos (x + \Delta x) \cos x}
= \displaystyle \lim_{\Delta x \rightarrow 0} \frac{1}{\cos (x + \Delta x) \cos x}
= \frac{1}{\cos x \cos x}
= \sec^2x

(b)

\frac{dy}{dx}
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{(x + \Delta x)e^{x + \Delta x} - xe^x}{\Delta x}
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{(x + \Delta x)e^xe^{\Delta x} - xe^x}{\Delta x}
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{e^x((x + \Delta x)e^{\Delta x} - x)}{\Delta x}
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{e^x(xe^{\Delta x} + \Delta xe^{\Delta x} - x)}{\Delta x}
= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{e^x(x(e^{\Delta x} - 1) + \Delta xe^{\Delta x})}{\Delta x}
= \displaystyle \lim_{\Delta x \rightarrow 0}e^x(x(\frac{e^{\Delta x} - 1}{\Delta x}) + e^{\Delta x})
= e^x(x(1) + 1)
= (x + 1)e^x

又或者用證明 product rule 那招,即考慮

(x + \Delta x)e^{x + \Delta x} - xe^x
= (x + \Delta x)e^{x + \Delta x} - xe^{x + \Delta x} + xe^{x + \Delta x} - xe^x

比較有系統地解之。

3 則迴響 »

  1. (a),(b) part入面都用了一些fact, e.g. lim sinx/x = 1 in (a)及l’hospital in (b)
    現在NSS也要學這些東西吧?

    題外話:john sir,我依家係教育局做緊intern,聽到D人開會討論緊邊D應該教,邊D唔應該,口氣上好像還未和考評局定好所有東西。。。

    迴響 由 Justin — 2010/08/01 @ 2:11 下午 | 回覆


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