[ 數學 ]differentiation

sky21212111

2009-08-24 6:34 am

1.let y = tan 1/x show that x^2 dy/dx + (y^2+1) = 0 .hence show that d^2y/dx^2 +2(x+y)/x^2 dy/dx = 0

2. let tanθ =2tanx and y = (tanθ - x ) where 0≦x≦π/2

a) express y in terms of tan x

b) where dy/dx = 0, find the value of x .

!!!!LONG STEP PLEASE!!!

回答 (1)

Prof. Physics

2009-08-24 7:05 am

✔ 最佳答案

1. y = tan(1/x)

dy/dx = -sec2(1/x) / x2

So, x2 dy/dx + (y2 + 1)

= x2[-sec2(1/x) / x2] + [tan2(1/x) + 1]

= -sec2(1/x) + sec2(1/x)

= 0

Differentiate both sides with respect to x

x2d2y/dx2 + 2xdy/dx + 2ydy/dx = 0

d2y/dx2 + 2(x + y)/x2 dy/dx = 0

2.a. y = tan(@ - x)

= [tan@ - tanx]/[1 + tan@tanx]

= (2tanx - tanx)/[1 + 2tanxtanx]

= tanx / (1 + 2tan2x)

dy/dx = (1 + 2tan2x)sec2x - tanx(4tanxsec2x) / (1 + 2tan2x)

= sec2x(1 - 2tan2x) / (1 + 2tan2x)

For dy/dx = 0

sec2x(1 - 2tan2x) = 0

sec2x =/= 0

So, 1 - 2tan2x = 0

tanx = 1/sqrt2 or -1/sqrt2 (rejected, since 0 <= x <= pi/2)

x = 0.615 (in radian)(cor. to 3 sig. fig.)

參考: Physics king

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