Question

Differentiate : log (1 + x²)  w.r.t. cot^-1 x. 

Answer 1

Let u =log (1 + x²) 

v = cot^-1 x 

To differentiate log ( 1 + x²) w.r.t. cot^-1 x is to find `"du"/"dv"`

Here u =log (1 + x²J and u = cot^-1 x,

where x is a parameter.

Differentiating w.r.t. x,

`"du"/"dx" = 1/(1 + "x"^2) xx "2x" ,  "dv"/"dx" = (-1)/(1 + "x"^2)`

`therefore "du"/"dv" = (("du"/"dx"))/(("dv"/"dx")) , "du"/"dx" != 0`

`= (("2x")/(1+"x"^2))/((-1)/(1 + "x"^2)) = "-2x"`