Quotient Rule, Chain Rule: [(x)(1 / {x^2 +1})(2x) - ln (x^2

[Lime]

How do you go from this:

[(x)(1 / {x^2 +1})(2x) - ln (x^2 + 1)] / x^2

To this:

[2x^2 - (x^2 + 1) ln(x^2 + 1)] / x^2(x^2 + 1)

Specifically I don't know how the (x^2 + 1) gets multiplied to the bottom, and I don't know where the ({1 / x^2 +1}) ends up.

 

[skeeter]

what is the original function you are working with to find the derivative?

 

[arthur ohlsten]

I have the wrong answer I must have interpreted your equation wrong


[x(1/{x^2+1} (2x) - ln(x^2+1) ] / x^2 rewrite
[2x^2 / {x^2+1} - ln (x^2+1] /x^2 rewrite as two terms
[2/{x^2+1}] - [ln(x^2+1]/x^2] rewrite
2 [x^2+1]^-1 - ln(x^2+1) / x^2

but derivative of :
d/dx 2[x^2+1]^-1 = -2[x^2+1]^-2 [2x]
d/dx 2[x^2+1]^-1= -4x/[x^2+1]^2 half the answer

but derivative ln[x^2+1] /x^2= x^2[2x]/[x^2+1]]-2xln[x^2+1] all over x^4
der ln[x^2+1] / x^2 = 2x^3/[x^4(x^2+1)] - 2x ln [x^2+1] / x^4
der ln ]x^2+1]= 2/[x(x^2+1)] - 2 ln[x^2+1] / x^3 other half of the answer
-4x/[x^2+1]^2 -2/[x(x^2+1)] + 2ln[x^2+1] /x^3

nothing like the expected answer I must have misinterpreted your equation

 

[Lime]

The original function is:

f(x) = [ln (x^2 + 1)] / x

Find f'(x).

 

[skeeter]

The original function is:

f(x) = [ln (x^2 + 1)] / x

Find f'(x).

\(\displaystyle \L f'(x) = \frac{x(\frac{2x}{x^2+1}) - ln(x^2+1)}{x^2}\)

\(\displaystyle \L f'(x) = \frac{\frac{2x^2}{x^2+1} - ln(x^2+1)}{x^2}\)

multiply numerator and denominator by (x² + 1) ...

\(\displaystyle \L f'(x) = \frac{2x^2 - (x^2+1)ln(x^2+1)}{x^2(x^2+1)}\)