Differentiation
Q. logx/x
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Answered by
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Answer :
Let, y = (logx)/x ...(i)
Now, differentiating both sides of (i), we get
dy/dx = d/dx (logx)/x
= {x d/dx (logx) - (logx) d/dx (x)}/(x^2)
= (x/x - logx)/(x^2)
= (1 - logx)/(x^2)
We have used the formula -
d/dx (u/v) = (v du/dx - u dv/dx)
Hope it helps!
Let, y = (logx)/x ...(i)
Now, differentiating both sides of (i), we get
dy/dx = d/dx (logx)/x
= {x d/dx (logx) - (logx) d/dx (x)}/(x^2)
= (x/x - logx)/(x^2)
= (1 - logx)/(x^2)
We have used the formula -
d/dx (u/v) = (v du/dx - u dv/dx)
Hope it helps!
Answered by
0
hii friend. ...
here is your answer. .
♥♥♥♥♥
★★★★★
y=logx/x
dy/dx=(x.d logx/dx-logx.dx/dx)/x^2
dy/dx=(x.1/x-logx)/x^2
=1-logx/x^2
.....
....
....
...
it is your answer
here is your answer. .
♥♥♥♥♥
★★★★★
y=logx/x
dy/dx=(x.d logx/dx-logx.dx/dx)/x^2
dy/dx=(x.1/x-logx)/x^2
=1-logx/x^2
.....
....
....
...
it is your answer
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