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:
C
Step-by-step explanation:
Line AC is equal to line DF.
Answer:
f=1
Step-by-step explanation:
Let start off with doing -f plus 4f which gives you 3f. The equation would now be 2+3f=8-3f. Add 3f so you can cancel out the -3f and add the 3f to the other 3f. Now your equation is 2+6f=8. Cancel out do by doing -2. (Don't forget to to it to the 8 also.) Now your equation is 6f=6. Divide 6 from both sides to get 1. (You divide because you want to get f alone.
Step-by-step explanation:
m
one set of ordered pair
b
It's A and C, I've taken a test that has this question and it was correct when I answered this, so A and C is your answer.
