100 POINTS
Evaluate the derivative of the function number 38

 1/ sqrt(1+ln^2(x))    * 1/(ln^2x +1) * 1/x

Step-by-step explanation:

f(x) = sin (tan^-1 (ln(x)))  

u substitution

       d/du (sin u)  * du /dx

            cos (u) * du/dx

Let u =(tan^-1 (ln(x)))     du/dx =d/dx  (tan^-1 (ln(x)))

v substitution

                                             Let v = ln x                dv/dx = 1/x

                                              d/dv  (tan ^-1 v)  dv/dx

                                                          1/( v^2+1) * dv/dx

                                                        =1/(ln^2x +1) * 1/x

Substituting this back in for  du/dx

cos (tan^-1 (ln(x))  * 1/(ln^2x +1) * 1/x

We know that cos (tan^-1 (a)) = 1/ sqrt(1+a^2)

cos (tan^-1 (ln(x))  * 1/(ln^2x +1) * 1/x

 1/ sqrt(1+ln^2(x))     * 1/(ln^2x +1) * 1/x

cos((tan^-1)(lnx))/[x(1 + ln²x)]

Step-by-step explanation:

Sin((tan^-1)(lnx))

First differentiate the outer function 'sin'

Then inner functions

derivative of:

sin(x) is cos(x)

tan^-(x) = 1/(1+x²)

lnx = 1/x

cos((tan^-1)(lnx)) × [1/(1 + ln²x)] × 1/x

cos((tan^-1)(lnx))/[x(1 + ln²x)]

17   x 256=4352

sorry if this is wrong, i'm new.

step-by-step explanation:

other answer is correct. i had used the radius of the radius somehow. a website: