Assume that y = x² ln x. Compute dy in terms of x and dx. Compute dy given that x = 1 and dx = 0.01?
Assume that y = x² ln x.
Compute dy in terms of x and dx.
Compute dy given that x = 1 and dx = 0.01
I tried deriving both sides do it was
dy = 2x * (1/x) dx
dy = 2 dx
so if dy = 2dx... why do we still need to be given the value of x?? i don't get it :(
cuz i would have plugged in dx = 0.01
so its 2 * 0.01
= 1/50.
Please tell me what i did wrong!
Update:Faz why did you add another +x at the end? Where did the +x in
2x (ln (x)) + x
come from?
thx
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y = x² ln x
dy/dx or derivative of y with respect to x is x² *(derivative of lnx)+(derivative of x^2)*lnx
dy/dx=x^2/x+2xlnx derivative of lnx=1/x
dy=dx(x+2xlnx) your first answeer<<<<
when x = 1 and dx = 0.01
dy=0.01(1+2*1(ln1))
dy=0.001+0.002ln(1)
ln(1)=0
dy=0.001
your answer
Happy to help
Basic deriving 101...
( F x G )' = F'G + G'F
And not F' x G'
so (x^2 ln x)' = 2x ln x + x^2/x = 2x ln x + x
dy/dx = 2x(ln(x)) + x
dy = ( 2x(ln(x)) + x )dx
Sub in x=1, dx=0.01 to get
(2(ln(1) + 1 ) 0.01 = 0.02