sin^2(cot^-1( (1+x)^0.5/ (1-x)^0.5))

Question

Please help me differentiates this. 

Answer 1

Notice that in one of the arguments, we are dividing square-root terms.  We can apply the rule of radicals to compact those arguments into one argument, thus simplifying the expression.

 

sin²[cot^-1[((1 + x)/(1 - x))^1/2]]

 

 

Note:  The argument of inverse cotangent is a square-root term, and you are taking the square-root of a quotient.

 

 

Let              w = cot^-1[((1 + x)/(1 - x))^1/2]

 

 

With this substitution, we get an easier expression to derive.

 

[sin(w)]²

 

 

To derive this simplified expression, use the chain rule.   By definition of the chain rule, the derivative of this expression is

 

2sin(w) * w'cos(w)

 

 

 

You will need to use this fact as well:

 

d/dx [cot^-1(x)] = -1 / (1 + x²)

 

 

This should help break it down.

Answer 2

Given  θ = cot^-1( (1+x)^0.5/ (1-x)^0.5) is an angle of a triangle.

 

Then cot(θ) = √(1 + x)/√(1 - x)

 

This is the cotangent of a right triangle and the legs of the triangle.

Draw a diagram to show how this is.

 

Let c = hypotenuse: C² = √(1 + x)² + √(1 - x)²
                                   = 1 + x + 1 - x = 2

The hypotenuse c = √(2)

 

From this triangle the sin(θ) = √(1 - x)/ √(2)

 

Now y = sin²(cot^-1( (1+x)^0.5/ (1-x)^0.5))

          = sin²(θ)

          = (√(1 - x)/ √(2))²

          = (1 - x)/2

 

Take the derivative, y' = -1/2 = -0.5

 

 

Answer 3

The derivative begins with 2sin(cot^-1(sqrt(1+x)/sqrt(1-x))) and then there follows the first chain-rule effect, a factor of 

-d/dx((sqrt(1+x)/sqrt(1-x))) ÷ [1+(1+x)/(1-x)].   Simplification is possible at this stage for the [ ] divisor.

 

Now you need to apply the quotient rule to get the derivative that I highlighted...use fractional exponents, factor out the lowest (-½) powers of (1-x) & (1+x), & simplify...

 

I got this final answer

-sin(cot^-1((sqrt(1+x)/sqrt(1-x)))•1/√(1-x²)   but  check everything carefully!!