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
2 answers
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
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
1 answer