calculus

prove that a constant makes the difference between function tan^-1 x and sin ^-1 x/squarert(1+x^2)

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Make a new function h(x) = arctan x-arcsin x/sqrt(1+x^2)

Differentiate the function h(x).

h'(x)=(arctan x-arcsin x/sqrt(1+x^2))'

Use chain rule for the derivative of arcsin x/sqrt(1+x^2)

h'(x)= 1/(1+x^2)-[1/sqrt(1-x^2/(1+x^2))]*(x/sqrt(1+x^2))'

Use product rule for the derivative of quotient x/sqrt(1+x^2)

(x/sqrt(1+x^2))'={(x)'*sqrt(1+x^2)-x*[sqrt(1+x^2)]'}/(1+x^2)

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

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

(x/sqrt(1+x^2))' = (x^2-x+1)/[(1+x^2)*sqrt(1+x^2)]

h'(x)= 1/(1+x^2)-[sqrt(1+x^2)]*{(1)/[(1+x^2)*sqrt(1+x^2)]}

h'(x) = 1/(1+x^2)-1/(1+x^2)

h'(x) = 0

The zero value of derivative prheoves that the function h(x) is a constant, such as h'(x)=0.

h(x)=c=arctan x-arcsin x/sqrt(1+x^2) => arctan x=c+arcsin x/sqrt(1+x^2)

**ANSWER:The difference between the two functions is made by the constant c.**

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