LOGARITHMIC FUNCTIONS, MATHS QUESTION. PLEASE HELPPP ME !!THANKYOU?

Question

Find the size of the obtuse angle ( to the nearest degree) between the two curves y=ln (x^2 +1) and y=2 ln (x+2) at their point of intersection.

Answer 1

The given two curves are y = ln (x² +1) and  y = 2 ln (x+2)

First find the point of intersection

ln (x² +1) = 2 ln (x+2)

ln (x² +1) = ln (x+2)²

(x² +1) = (x+2)²

x² +1 = x² + 4x + 4

4x + 4 = 1

4x = 1 – 4

The slope at x = –3/4

y = ln ((–3/4)² +1) = 0.4462871

y = 2 ln ((–3/4)+2) = 0.4462871

Now find the tangent to each line at that point
line 1

y = ln (x² +1)

chain rule
If y = f(u) and u = g(x)

dy/dx = (dy/du)(du/dx)

u = (x² +1)

du/dx = 2x

y = ln u

dy/du = 1/u

dy/dx = (1/(x² +1))(2x)

y' = 2x / (x² +1)

The slope at x = –3/4

y' = 2(–3/4) / ((–3/4)² +1)

  = (–3/2) / ((9/16)+1) = –(3/2) / (25/16)

y' = –(3/2) (16/25) = –(3) (8/25) = –24/25

Now find the tangent to each line at that point
line2

y = y = 2 ln (x+2)

 Apply chain rule
If y = f(u) and u = g(x)

dy/dx = (dy/du)(du/dx

u = (x+2)²

du/dx = 2 (x+2)

y = ln u

dy/du = 1/u

dy/dx = (1/u) 2(x+2)

y' = (1/(x+2)² ) ( 2(x+2))

y' = 2/(x+2)

The slope at x = –3/4

y' = 2/( -3/4)+2

= 2( 4/5) = 8/5

Use the two slopes to get the angle tanθ =( m1 - m2)/1+m1m2

m1 = -24/25 and m2 = 8/5

tanθ = [( -24/25 ) - (8/5)] /1+ (-24/25)(8/5)

=[( -24 - 40)/25]/[( 125 - 192)/125]

= ( -64/-67) (5) = 320/67

θ = tan-1(320/67)

θ = 78.2 degrees

The angle between two slopes is θ = 78.2 degrees.