Answer:
hard, search on internet as I don't know this
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:
It would be in the $3.30 or lower zone
Step-by-step explanation:
Answer:
$6
Step-by-step explanation:
Hourly rate is $1.5
