To whom may know:
Differentiate the following function
y= X ^3x
Thank you very much!!
Differentiate the following function
2007-01-11 7:37 am
回答 (3)
2007-01-11 7:47 am
y=x^3x
ln y = 3x ln x
(1/y)dy/dx =3 ln x +3 x(1/x)=3 (ln x + 1)
dy/dx=3x^3x (ln x +1) (ANS.)
ln y = 3x ln x
(1/y)dy/dx =3 ln x +3 x(1/x)=3 (ln x + 1)
dy/dx=3x^3x (ln x +1) (ANS.)
2007-01-11 7:46 am
y= x^3x
ln y=3x(lnx)
so (1/y)y'=3x/x+3lnx
so (1/y)y'=3+3lnx
y'=3(1+lnx)*y
=3(ln+1)(x^3x)
ln y=3x(lnx)
so (1/y)y'=3x/x+3lnx
so (1/y)y'=3+3lnx
y'=3(1+lnx)*y
=3(ln+1)(x^3x)
2007-01-11 7:45 am
∵dlny/dx=dlny/dy*dy/dx (chain rule)
∴dy/dx = (dlny/dx) / (dlny/dy)
ln y = 3x ln x
∴dy/dx = (d3xlnx/dx) / (dlny/dy)
= (lnx d3x/dx + 3x dlnx/x) / (1/y)
= (3lnx + 3x*1/x) / (1/y)
= 3y(lnx + 1)
=3(x^3x)(lnx+1)
∴dy/dx = (dlny/dx) / (dlny/dy)
ln y = 3x ln x
∴dy/dx = (d3xlnx/dx) / (dlny/dy)
= (lnx d3x/dx + 3x dlnx/x) / (1/y)
= (3lnx + 3x*1/x) / (1/y)
= 3y(lnx + 1)
=3(x^3x)(lnx+1)
參考: me
收錄日期: 2021-04-12 21:22:13
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