Find the Derivative - d/dx y=arctan( square root of (1+x)/(1-x))

Problem

d()/d(x)*arctan(√(,(1+x)/(1−x)))

Solution

1. Identify the outer function and the inner function to apply the Chain Rule. The outer function is arctan(u) and the inner function is u=√(,(1+x)/(1−x))

2. Apply the derivative formula for the arctangent function, which is d(arctan(u))/d(x)=1/(1+u2)⋅d(u)/d(x)

3. Substitute the expression for u into the first part of the derivative.

1/(1+(√(,(1+x)/(1−x)))2)=1/(1+(1+x)/(1−x))

4. Simplify the denominator of the first part by finding a common denominator.

1+(1+x)/(1−x)=(1−x+1+x)/(1−x)=2/(1−x)

1/2/(1−x)=(1−x)/2

5. Differentiate the inner function u=((1+x)/(1−x))(1/2) using the Power Rule and the Quotient Rule.

d(u)/d(x)=1/2*((1+x)/(1−x))(−1/2)⋅d()/d(x)(1+x)/(1−x)

6. Apply the Quotient Rule to the innermost fraction.

d()/d(x)(1+x)/(1−x)=((1)*(1−x)−(1+x)*(−1))/((1−x)2)

(1−x+1+x)/((1−x)2)=2/((1−x)2)

7. Combine the components of d(u)/d(x)

d(u)/d(x)=1/2√(,(1−x)/(1+x))⋅2/((1−x)2)

d(u)/d(x)=√(,1−x)/(√(,1+x)*(1−x)2)

8. Multiply the simplified outer derivative by the inner derivative.

d(y)/d(x)=(1−x)/2⋅√(,1−x)/(√(,1+x)*(1−x)2)

9. Simplify the final expression by canceling terms.

d(y)/d(x)=√(,1−x)/(2√(,1+x)*(1−x))

d(y)/d(x)=1/(2√(,1+x)√(,1−x))

d(y)/d(x)=1/(2√(,1−x2))

Final Answer

d()/d(x)*arctan(√(,(1+x)/(1−x)))=1/(2√(,1−x2))

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