I have gotten so many different responses from mine on this question
Can someone PLEASE write out the steps for me for this. Thanks
f'(x)= 2x/(x^2+1)
Find the second derivative.
I get 4x^3 + x^2- 12x + 1
Oops I meant 4x^3 + x^2 - 12x + 1/ ((x^2+1)^3)
how can you possibly get a polynomial as the derivative of a rational function?
apparently
f(x) = ln(x^2+1)
f'(x) = 2x/(x^2+1)
f"(x) = 2(1-x^2)/(x^2+1)^2
since if
f =u/v
f' = (u'v-uv')/v^2
so, the derivative of f' is
((2)(x^2+1) - (2x)(2x))/(x^2+1)^2
= (2x^2+2 - 4x^2)/(x^2+1)^2
= 2(1-x^2)/(x^2+1)^2
Now, if you want one more derivative, as you apparently do,
[2(-2x)(x^2+1)^2 - 2(1-x^2)2(2x)(x^2+1)]/(x^2+1)^4
= (4x^5 - 8x^3 - 12x)/(x^2+1)^4
= (4x(x^2-3)(x^2+1))/(x^2+1)^4
= 4x(x^2-3)/(x^2+1)^3
= 4x(x^2-1)/(x^2+1)^3
To find the second derivative of the function f(x), which is represented by f'(x) = 2x/(x^2+1), you need to apply the derivative operation twice.
Step 1: Find the first derivative of f(x)
To find the first derivative, you can use the quotient rule. The quotient rule states that for a function h(x) = g(x)/k(x), where g(x) and k(x) are differentiable functions, the derivative of h(x) is given by:
h'(x) = (g'(x) * k(x) - g(x) * k'(x)) / (k(x))^2
Let's apply the quotient rule to f(x) = 2x/(x^2+1):
f'(x) = ((2 * (x^2+1)) - (2x * 2x))/(x^2+1)^2
= (2x^2 + 2 - 4x^2)/(x^2+1)^2
= (-2x^2 + 2)/(x^2+1)^2
So, the first derivative of f(x) is f'(x) = (-2x^2 + 2)/(x^2+1)^2.
Step 2: Find the second derivative of f(x)
To find the second derivative, you need to differentiate f'(x) with respect to x. Apply the quotient rule again:
f''(x) = [(-2 * 2x * (x^2+1)^2) - ((-2x^2 + 2) * (2 * (x^2+1) * 2x))]/((x^2+1)^2)^2
= [(-4x(x^2+1)^2) - (4x(-2x^3-2x+x^2+1))]/((x^2+1)^2)^2
= [(-4x(x^2+1)^2) + (8x^4 + 4x^3 - 4x^2 - 4x)]/((x^2+1)^2)^2
= (-4x^5 - 8x^4 + 4x^3 - 4x^2 - 4x)/(x^2+1)^4
Thus, the second derivative of f(x) is f''(x) = (-4x^5 - 8x^4 + 4x^3 - 4x^2 - 4x)/(x^2+1)^4.
Hence, the second derivative is given by f''(x) = -4x^5 - 8x^4 + 4x^3 - 4x^2 - 4x divided by (x^2+1)^4.