Finding derivatives

Question

Find the derivative of the function.

 

y = arctan(square root of (1-x)/1+x))

 

 

I could not add the square root symbol 

Answer 1

The derivative of arctan(u) = 1/(1+u²) • du/dx.

The derivative of u^1/2 = (1/2)u^-1/2•du/dx

The derivative of f(x)/g(x) = [g(x)f'(x) - f(x)g'(x)]/[g(x)²]

Applying those formulas:

y ' = 1/[1 + (1-x)/(1+x)] • (1/2)[(1-x)/(1+x)]^-1/2 • [(1+x)(-1) - (1-x)(1)]/(1+x)²

Simplifying each of the above factors:

For the first factor multiply numerator and denominator by (1+x) to clear the complex fraction, resulting in:

(1+x)/[(1+x) +(1-x)] = (1+x)/2

For the 2nd factor rewrite using a square root symbol:

1/2√[(1-x)/(1+x)]

The 3rd factor:

-2/(1+x)²

y' = (1+x)/2 • 1/2√[(1-x)(1+x)] • -2/(1+x)²

One factor of 2 cancels in the denominator. One factor of (1+x) also cancels:

y' = -1 /2 [√[(1-x)(1+x)] (1+x)]

If you introduce the (1+x) into the square root:

y' = -1 / 2√(1 - x²)

If you choose to rationalize the denominator:

y' = -√(1-x²)/2(1-x²)

desmos.com/calculator/skcrdjubug

The above graph shows that F(x) [the original function] has domain -1 ≤ x ≤1 and that F(x) has vertical tangent lines at x = -1 and x = 1.