What is the second derivative of $f(x)=(x-1)/(x^2+1)$ ?

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Answer 1 · 2017-03-06T14:54:08

$f''(x)={2(1+x)(1-4x+x^2)}/(1+x^2)^3.$

$f(x)=(x-1)/(x^2+1).$

Using the Quotient Rule for Diffn., we get, the first deri.,

$f'(x)={(x^2+1)d/dx(x-1)-(x-1)d/dx(x^2+1)}/(x^2+1)^2$

$={(x^2+1)(1)-(x-1)(2x)}/(x^2+1)^2={x^2+1-2x^2+2x}/(x^2+1)^2$

$:. f'(x)=(1+2x-x^2)/(1+2x^2+x^4)$

Rediff.ing $f'(x)$ to get second deri.,

$f''(x)={(1+x^2)^2d/dx(1+2x-x^2)-(1+2x-x^2)d/dx(1+2x^2+x^4)}/{(x^2+1)^2}^2$

$={(1+x^2)^2(2-2x)-(1+2x-x^2)(4x+4x^3)}/(x^2+1)^4$

$={2(1+x^2)^2(1-x)-4x(1+x^2)(1+2x-x^2)}/(x^2+1)^4$

$=[2(1+x^2){(1+x^2)(1-x)-2x(1+2x-x^2)}]/(x^2+1)^4$

$=[2(1+x^2){1+x^2-x-x^3-2x-4x^2+2x^3}]/(x^2+1)^4$

$:. f''(x)={2(1-3x-3x^2+x^3)}/(1+x^2)^3.$

$=[2{(1+x^3)-3x(1+x)}]/(1+x^2)^3$

$=[2{(1+x)(1-x+x^2)-3x(1+x)}]/(1+x^2)^3$

$=[2(1+x)(1-x+x^2-3x)}/(1+x^2)^3$

$rArr f''(x)={2(1+x)(1-4x+x^2)}/(1+x^2)^3$

Enjoy Maths.!

What is the second derivative of f(x)=(x-1)/(x^2+1)f(x)=(x-1)/(x^2+1)?