Derivative: $f(x)= ln \sqrt { x^2 -+1 } } {x} {2x^3-1}^2\]

[venialove]

Find the derivative: $f(x)= ln \sqrt { x^2 -+1 } } {x} {2x^3-1}^2\]

or f(x) =ln ?(x^2+1) divided by x(2x^3-1)^3

 

[Deleted member 4993]

Re: Derivative

Find the derivative: $f(x)= ln \sqrt { x^2 -+1 } } {x} {2x^3-1}^2\]

or f(x) =ln ?(x^2+1) divided by x(2x^3-1)^3

You will need to apply chain rule and product rule and exponent rule.

Please show us your work - indicating exactly where you are stuck.

 

[venialove]

Re: Derivative

f(x)= ln ? (x^2+1) divided by x(2x^3-1)^2

= ln x + 2 (2x divided by x^2+ 1) - 1/2 ln ( 2x^3-1)

f''(x)= 1/x +2 ( 2x divided by x^2+ 1) minus 1/2 (6x^2 divided by 2x^3-1)

 

[Deleted member 4993]

Re: Derivative

f(x)= ln ? (x^2+1) divided by x(2x^3-1)^2

= ln x + 2 (2x divided by x^2+ 1) - 1/2 ln ( 2x^3-1)

f''(x)= 1/x +2 ( 2x divided by x^2+ 1) minus 1/2 (6x^2 divided by 2x^3-1)

Is your problem:

find f'(x) when

\(\displaystyle f(x)\, = \, ln\sqrt{\frac{x^2 \, + \, \, 1}{x \cdot (2x^3-1)}\)

or

\(\displaystyle f(x)\, = \, \frac{ln\sqrt{x^2 \, + \, \, 1}}{x \cdot (2x^3-1)}\)

or

\(\displaystyle f(x)\, = \, ln\frac{\sqrt{x^2 \, + \, \, 1}}{x \cdot (2x^3-1)}\)

or something else?

 

[venialove]

It's like the last one but the bottom part has a exponent of 2 like this
x(2x^3-1)^2

 

[Deleted member 4993]

Re: Derivative

f(x)= ln ? (x^2+1) divided by x(2x^3-1)^2

= ln x + 2 (2x divided by x^2+ 1) - 1/2 ln ( 2x^3-1)

f''(x)= 1/x +2 ( 2x divided by x^2+ 1) minus 1/2 (6x^2 divided by 2x^3-1)

your problem:

find f'(x) when

\(\displaystyle f(x)\, = \, ln\frac{\sqrt{x^2 \, + \, \, 1}}{x \cdot (2x^3-1)^2}\)

\(\displaystyle f(x)\, = \, \frac{1}{2}\cdot ln[x^2 \, + \, \, 1] \, - ln[x] \, - \, 2\cdot ln(2x^3-1)\)

\(\displaystyle f'(x)\, = \, \frac{1}{2}\cdot \frac{1}{x^2 \, + \, \, 1}\cdot [2x] \, - \frac{1}{[x]} \, - \, 2\cdot \frac{1}{2x^3-1}\cdot (6x^2)\)

Now simplify and finish.....

or something else?