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What is the f''(x) of (x/(x^{2}+2))

I tried using the calculato by web 2.0 but I think it's incorrect.

Guest Aug 5, 2017

edited by
Guest
Aug 5, 2017

#1**0 **

Possible derivation:

d/dx((x f''(x))/(2 + x^2))

Use the quotient rule, d/dx(u/v) = (v ( du)/( dx) - u ( dv)/( dx))/v^2, where u = x f''(x) and v = x^2 + 2:

= ((2 + x^2) (d/dx(x f''(x))) - x (d/dx(2 + x^2)) f''(x))/(2 + x^2)^2

Use the product rule, d/dx(u v) = v ( du)/( dx) + u ( dv)/( dx), where u = x and v = f''(x):

= ((2 + x^2) x (d/dx(f''(x))) + (d/dx(x)) f''(x) - x (d/dx(2 + x^2)) f''(x))/(2 + x^2)^2

The derivative of f''(x) is f^(3)(x):

= (-x (d/dx(2 + x^2)) f''(x) + (2 + x^2) (f^(3)(x) x + (d/dx(x)) f''(x)))/(2 + x^2)^2

Differentiate the sum term by term:

= (-x f''(x) d/dx(2) + d/dx(x^2) + (2 + x^2) ((d/dx(x)) f''(x) + x f^(3)(x)))/(2 + x^2)^2

The derivative of 2 is zero:

= (-x (d/dx(x^2) + 0) f''(x) + (2 + x^2) ((d/dx(x)) f''(x) + x f^(3)(x)))/(2 + x^2)^2

Simplify the expression:

= (-x (d/dx(x^2)) f''(x) + (2 + x^2) ((d/dx(x)) f''(x) + x f^(3)(x)))/(2 + x^2)^2

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 2: d/dx(x^2) = 2 x:

= (-x f''(x) 2 x + (2 + x^2) ((d/dx(x)) f''(x) + x f^(3)(x)))/(2 + x^2)^2

Simplify the expression:

= (-2 x^2 f''(x) + (2 + x^2) ((d/dx(x)) f''(x) + x f^(3)(x)))/(2 + x^2)^2

The derivative of x is 1:

**Answer: | = (-2 x^2 f''(x) + (2 + x^2) (1 f''(x) + x f^(3)(x)))/(2 + x^2)^2**

Guest Aug 5, 2017