I am confused on how to go about the following calculus questions

Question

Solve:

1.) √ ln x = ln √ x

2.) Show that d/dx (1/2 tan −1 x + 1/4 ln (x + 1)2 /x 2 + 1 ) = 1 / (1 + x) (1 + x 2)

3.) At what point on the curve y = [ln (x + 4)]2 is the tangent horizontal?

Answer 1

1)

√ln(x) = ln(√x)

√ln(x) = ln(x)/2 by rewriting ln(x^1/2)

ln(x) = ln²(x)/4 by squaring both sides

4ln(x) - ln²(x) = 0 by multiplying by 4 and bringing to the same side

ln(x) (4 - ln(x)) = 0 by factoring ln(x)

Solution 1:

ln(x) = 0

x = 1

Solution 2:

4 - ln(x) = 0

x = e⁴

2)

Just for clarity lets write

f(x) = tan^-1(x)

g(x) = ln((x+1)²/(x²+1)) = ln(x+1)² - ln(x²+1) = 2 ln(x+1) - ln(x²+1)

You need to evaluate

d/dx(f(x)/2 + g(x)/4) = f'(x)/2 + g'(x)/4

f'(x) = 1/(x²+1)

g'(x) = 2/(x+1) - 2x/(x²+1)

Now for the final answer

f'(x)/2 + g'(x)/4 = 1/2(x²+1) + 1/2(x+1) - x/2(x²+1)

= (x+1 + x²+1 - x(x+1))/2(x+1)(x²+1)

= 1/(x+1)(x²+1)

3) The tangent is horizontal at for those x where the derivative is 0

y' = 2 ln(x+4) / (x+4) = 0

ln(x+4) = 0

x+4 = 1

x = -3

Answer 2

I can help you with parts1 and 3. Part 2 is just a messy differentiation problem.

Part one: graph the difference between the RHS and the LHS...you will see immediately that there is only one root, namely x=1.

Part 3: the derivative is 2ln(x+4)/(x+4) which is zero when x+4=1, i.e. x=-3