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प्रश्न
Differentiate : log (1 + x2) w.r.t. cot-1 x.
उत्तर
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"`
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