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
subtract 1/3x
so that you can get qui equatio
Answer:
It is shifted left 5 units and up 2 units from the parent function.
Step-by-step explanation:
Given
Required
Compare both functions
First, translate y, 5 units left.
The rule is:
So, we have:
Next, translate y1, 2 units up.
The rule is:
So, we have:
Hence, the transformation is:
5 units left and 2 units up
Answer:
A linear pair is a pair of angles that share a side and a base. In other words, they are the two angles created along one line when two lines intersect. Linear pairs are always supplementary.
