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
The correct answer should be 7 levels for the entirety of the week. However, there are two solutions that satisfy this question so perhaps it is really asking how many additional levels he is able to beat. In this case, the answer should be B: x > 5 2/3
Answer:
d=3 and e=-1
Step-by-step explanation:
d+e=2
d=2-e-----(1)
d-e=4-----(2)
substituting (1) in (2)
2-e-e=4
-2e=2
e=-1-----(3)
substituting (3) in (1)
d=2-(-1)
d=3
Answer:
g(f(x)) = -6x+6.
Step-by-step explanation:
It is given that,
f(x) = -3x + 2 and g(x) = 2x – 5
We need to find g(f(x)).
Put it into g(x).
g(f(x)) = 2(-3x + 2)+2
= -6x+4+2
=-6x+6
Hence, the value of g(f(x)) is -6x+6.
