Differentiate y=4x+1/x?
回答 (20)
2016-06-25 4:56 pm
Case I : y = (4x + 1)/x
y' = [x(4x - 1)' - (4x - 1)x'] / x²
= [x(4) - (4x - 1)] / x²
= (4x - 4x + 1) / x²
= 1/x²
====
Case II : y = 4x + (1/x)
y' = (4x)' + (1/x)'
= 4 + (-1/x²)
= 4 - (1/x²)
y' = [x(4x - 1)' - (4x - 1)x'] / x²
= [x(4) - (4x - 1)] / x²
= (4x - 4x + 1) / x²
= 1/x²
====
Case II : y = 4x + (1/x)
y' = (4x)' + (1/x)'
= 4 + (-1/x²)
= 4 - (1/x²)
2016-06-29 5:22 pm
Do you mean 4x + (1/x), or (4x+1)/x ?
2016-06-26 9:30 pm
4 - 1/x²
2016-06-25 4:54 pm
dy/dx = 4 - 1/x²
2016-07-03 2:46 am
dy/dx = 4 - 1/x^2 for x <> 0
2016-07-02 11:12 am
dy/dx = 4 - 1/x^2
2016-06-28 6:31 am
Answer y =4x+1/x
dy/dx = d/dx(4x+1/x)
dy/dx = 4*1-1/x²
dy/dx = 4-1/x²
dy/dx = d/dx(4x+1/x)
dy/dx = 4*1-1/x²
dy/dx = 4-1/x²
2016-06-28 5:57 am
y'(x) = 4 - 1/x^2
2016-06-25 6:42 pm
dy/dx = 4 - 1/x^2
2016-06-25 4:55 pm
Well,
y = 4x + 1/x
then
dy/dx = 4 - 1/x^2
now, if you mean
y = (4x + 1)/x <---- then better PUT BRACKETS !! ;-)
dy/dx = ( 4x - (4x+1) ) /x^2
= -1/x^2
hope it' ll help !!
y = 4x + 1/x
then
dy/dx = 4 - 1/x^2
now, if you mean
y = (4x + 1)/x <---- then better PUT BRACKETS !! ;-)
dy/dx = ( 4x - (4x+1) ) /x^2
= -1/x^2
hope it' ll help !!
2016-07-05 2:33 pm
dy/dx=d(4x+1/x)/dx
D(y)= D(4x)+D(1/x)
D(y)=4+(-1/x^2)
D(y)=4-1/x^2
D(y)= D(4x)+D(1/x)
D(y)=4+(-1/x^2)
D(y)=4-1/x^2
2016-07-05 6:15 am
dy/dx = 4 - 1/x^2
2016-07-02 5:25 pm
dy/dx=4-1/x^2
2016-07-02 7:52 am
4 - x^-2
2016-06-26 10:05 pm
y = 4x + x^(-1), dy/dx = 4 - (1/x^2)
y = (4x+1)/x = 4x/x + 1/x = 4+1/x , dy/dx = -1/x^2
y = (4x+1)/x = 4x/x + 1/x = 4+1/x , dy/dx = -1/x^2
2016-06-26 1:41 pm
y = 4x + x^(-1)
y' = 4 + (-1) x^(-2)
y' = 4 - 1 / x^2
y' = 4 + (-1) x^(-2)
y' = 4 - 1 / x^2
2016-06-25 4:56 pm
dy/dx = 4 - 1/x²
2016-07-01 9:44 pm
y = 4x + 1/x
y = 4x + x^-1
dy/dx = 4 + (-1)(x^-2)
dy/dx = 4 - x^-2
dy/dx = 4 - 1/x^2
y = 4x + x^-1
dy/dx = 4 + (-1)(x^-2)
dy/dx = 4 - x^-2
dy/dx = 4 - 1/x^2
2016-06-25 4:54 pm
y= 4x+1/x
y= 4+1/x
y'= 0-1/x2
y'= -1/x2
y= 4+1/x
y'= 0-1/x2
y'= -1/x2
2016-06-26 9:20 am
-1/x^2
收錄日期: 2021-04-18 15:11:16
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