Question

find dy/dx of y=ln(X/1+x^2)

Answer 1

the derivative of
ln(u)=\[\frac{ 1 }{ u }*\frac{ d }{ du }(u)\]

Answer 2

can you use this rule?

Answer 3

Yeah im just having trouble finding the u

Answer 4

in your problem
u=\[\frac{ x }{ 1+x ^{2} }\]

Answer 5

then the ans will be
\[\frac{ 1 }{ \frac{ x }{ 1+x ^{2} } }*\frac{ d }{ du }(\frac{ x }{ 1+x ^{2} })\]

Answer 6

So I take the derivative of the u with a quotient rule?

Answer 7

yes right

Answer 8

or before you derive it, by using the property of log it can be
y = ln(x/1+x^2) = ln(x) - ln(1+x^2)
now you can derive it one by one