Answer:
1/ sqrt(1+ln^2(x)) * 1/(ln^2x +1) * 1/x
Step-by-step explanation:
f(x) = sin (tan^-1 (ln(x)))
u substitution
d/du (sin u) * du /dx
cos (u) * du/dx
Let u =(tan^-1 (ln(x))) du/dx =d/dx (tan^-1 (ln(x)))
v substitution
Let v = ln x dv/dx = 1/x
d/dv (tan ^-1 v) dv/dx
1/( v^2+1) * dv/dx
=1/(ln^2x +1) * 1/x
Substituting this back in for du/dx
cos (tan^-1 (ln(x)) * 1/(ln^2x +1) * 1/x
We know that cos (tan^-1 (a)) = 1/ sqrt(1+a^2)
cos (tan^-1 (ln(x)) * 1/(ln^2x +1) * 1/x
1/ sqrt(1+ln^2(x)) * 1/(ln^2x +1) * 1/x
Answer:
Khannas salary is $547
Step-by-step explanation
Literally all you have to do is subtract 584 by 37.
The idea is to use the tangent line to 
at 
in order to approximate 
.
We have
so the linear approximation to 
is
Hence 
and 
.
Then
The graph of y = (-3x)^2 is much narrower than it's original graph, but still keeps all of the other properties of it's parent parable y = x^2. The new graph's with is that of the original parabola's with divided by 3.
