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differentiation

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1. f(3)=1,f'(3)=-4,g(3)=-1,g'(3)=3calculate(f+g)'(3)=?(f-g)'(3)=?(fxg)'(3)=?(f/g)'(3)=?2. if d(f(4x^2))/dx=9x^3,find f'(x)3.If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 32 ft/sec, its height after t seconds is... 顯示更多 1. f(3)=1,f'(3)=-4,g(3)=-1,g'(3)=3 calculate (f+g)'(3)=? (f-g)'(3)=? (fxg)'(3)=? (f/g)'(3)=? 2. if d(f(4x^2))/dx=9x^3,find f'(x) 3.If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 32 ft/sec, its height after t seconds is s(t)=64+32t-16t^2 a.) What is the maximum height the ball reaches? b.) What is the velocity of the ball when it hits the ground (height0 )?

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1. f(3)=1,f'(3)=-4,g(3)=-1,g'(3)=3 calculate (f+g)'(3)=? (f-g)'(3)=? (fxg)'(3)=? (f/g)'(3)=? d(f + g)/dx = df/dx + dg/dx (f+g)' = f' + g' (f+g)'(3) = f'(3) + g'(3) = -4 + 3 = -1 d(f - g)/dx = df/dx - dg/dx (f-g)' = f' - g' (f-g)'(3) = f'(3) - g'(3) = -4 - 3 = -7 d(fg)/dx = f(dg/dx) + g(df/dx) (fxg)' = fg' + gf' (fxg)'(3) = f(3)g'(3) + g(3)f'(3) = (1)(3) + (-1)(-4) = 3 + 4 = 7 2013-10-04 22:51:11 補充: 其實,這是 +-×÷ 的公式的代入,初時對 df/dx 和 f' 之類的寫法不習慣,會感到困難,但你看完我為你做的 3 條之後,應該在感覺上會好了一些,自己試除式啦!它是很複雜的,小心去做,做唔到的話,再問。 2013-10-04 23:05:07 補充: df(4x^2)/dx = 9x^3 設 y = 4x^2 則有 df(y)/dx = 9x^3 [df(y)/dy][dy/dx] = 9x^3 [df(y)/dy][8x] = 9x^3 [df(y)/dy] = (9/8)x^2 df(y)/dy = (9/8)(4x^2/4) = (9/8)(y/4) = (9/32)y y 和 x 都只是符號一個,所以 df(x)/dx = (9/32)x 2013-10-04 23:10:00 補充: s(t)=64+32t-16t^2 s'(t) = 32 - 32t s''(t) = -32 ; s''(t)<0 ; that means when s'(t) = 0 is a max. when s'(t) = 0 ==> 32 - 32t = 0 ; t = 1 max. height = s(1) = 64 + 32(1) - 16(1)^2 = 64 + 32 - 16 = 80 ft 2013-10-04 23:21:16 補充: velocity = s'(t) = 32 - 32t ; when the ball hits the ground s(t) = 0 0=64+32t-16t^2 16t^2 - 32t - 64 = 0 t^2 - 2t - 4 = 0 t = [2±√(4+16)]/2 t = 1±√5 = 3.236 or -1.236(reject) 代入 velocity = s'(t) = 32 - 32t 便完成了。 = -71.552 ft/s。 這是一個負數,它是正確的,負數代表什麼意思呢?想想!想想!

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