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# 微積分計算問題

<1>suppose that a differentiable function y=f(x) is implicitly defined by

the equation x^y=y^x .

find dy/dx, dx/dy

<2>Estimate∫sin(t^2) dt correct to within an error 0.001

範圍是0~1

<3>∫sec^5x dx

dx/dy =x ( x - lnx ) /y( y - x lny )

ln x前面是不是少ㄍy

### 1 Answer

- densityLv 61 decade agoFavorite Answer
1.xy = yx, =>ln xy = ln yx so that y lnx = x lny . Differentiate on both sides with respect to x , we get y'．( lnx )+ y．( 1 / x )= 1．ln y+ x．(y'/ y). And hence dy/dx = y' =y( y - x lny )/x ( x - lnx ), where dx/dy =1/(dy/dx) ∴dx/dy =x ( x - lnx ) /y( y - x lny ) . 2.Estimate∫10 sin( t2 )dt correct to within an error 0.001. (See my blog! )3.∫sec5 xdx We use the reduced form to find the answer. Let In=∫secn xdx=1/(n-1)．secn-2 x．tanx+(n-2/n-1)．∫secn-2 xdx , thus ∫sec5 xdx =(1/4 )．sec3 x．tanx+(3/4)．I 3 =(1/4 )．sec3 x．tanx+(3/4)．((1/2)secx．tanx + (1/2)∫secxdx )==(1/4 )．sec3 x．tanx+(3/8)．secx．tanx +(3/8)．ln|secx+tanx|+c

2008-02-02 06:49:49 補充：

dx/dy =x ( x -y lnx ) /y( y - x lny )

漏打了!!

Source(s): myself