Question

Take the derivative of v(x)=x^1f(x)+\frac{g(x) }{ x^2 }

Answer 1

\[v(x)=x^1f(x)+\frac{ g(x) }{ x^2 }\]

Answer 2

If \(w(x) = u(x)v(x)\), then \(w'(x) = u(x)v'(x) + v(x)u'(x)\).

Answer 3

So it would be like this:
\[v'(x)= ((x^1)'f(x)+(x^1)f'(x))+(\frac{ g(x)'x^2-g(x)(x^2)' }{ (x^2)^2 })\]
@mathstudent55

Answer 4

Yes, looks good.
You should ignore the arguments to make it look a little nicer lol\[\large\rm v=x'f+xf'+\frac{g'x^2-g(x^2)'}{(x^2)^2}\]