Question

What will be the differentiation of it?

Answer 1

`d/dx arctan((sqrt(1-x^2)+1)/x^2 ) = -(1/(xsqrt(1-x^2)) + (2(sqrt(1-x^2)+1)) / x^3)/(1+((sqrt(1-x^2)+1)/x^2 )^2)`

Explanation:

We seek:

`d/dx arctan((sqrt(1-x^2)+1)/x^2 )`

For simplicity, let:

`y = arctan(( sqrt(1-x^2)+1)/x^2 )`

Theen, using the result:

`d/dx arctan x = 1/(1+x^2)`

In conjuction with the chain rule, we have:

`dy/dx = 1/(1+(sqrt(1-x^2+1)/x^2 )^2) \ d/dx ((sqrt(1-x^2)+1)/x^2 )`

Then we apply the quotient rule, to get:

`dy/dx = 1/(1+((sqrt(1-x^2)+1)/x^2 )^2) \ ( (x^2)(d/dx sqrt(1-x^2)+1) - (sqrt(1-x^2)+1)(d/dx x^2) ) / (x^2)^2`

`\ \ \ \ \ \ = 1/(1+((sqrt(1-x^2)+1)/x^2 )^2) \ (x^2 1/2(1-x^2)^(-1/2)(-2x) - (sqrt(1-x^2)+1)2x ) / (x^4)`

`\ \ \ \ \ \ = 1/(1+((sqrt(1-x^2)+1)/x^2 )^2) \ (-x^3 (1-x^2)^(-1/2) - (sqrt(1-x^2)+1)2x ) / (x^4)`

`\ \ \ \ \ \ = -(1/(xsqrt(1-x^2)) + (2(sqrt(1-x^2)+1)) / x^3)/(1+((sqrt(1-x^2)+1)/x^2 )^2)`

Further simplification is possible and is an exercise in Algebra rather than Calculus, we find that:

`dy/dx = -(x(2x^2+sqrt(1-x^2)(x^2-3)-2))/(x^6-3x^4+7x^2-5)`