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
y-intercept is -6 (0,-6) and the x-intercept is -3 (-3,0)
Answer:
x=1
Step-by-step explanation:
x = -4 + 6 + (11 + 4(-3)).
First get the numbers in the parentheses
x = -4 + 6 + (11 - 12)
Since 11 - 12 = -1. We can replace (11-12) with -1
x = -4 + 6 -1
Now add and subtract
x = 1
