can someone help me with calculus please?? thank u!!?

Question

how do i differentiate this?
y=3x/x^2+1

Answer 1

y=3x/x^2+1

Take log of both sides w.r.t. base 'e'
ln(y) = ln{3x/(x^2 + 1)}
ln(y) = ln(3) + ln(x) - ln(x^2 + 1)

Differentiating both sides w.r.t. x
(1/y)(dy/dx) = 0 + 1/x - 2x/(x^2 + 1)
(1/y)(dy/dx) = (x^2 - 2x^2 + 1)/{x(x^2 + 1)}
dy/dx = -y(x^2 - 1) / {x(x^2 + 1)}
dy/dx = -3x(x^2 - 1) / {x(x^2 + 1)^2} [substituting for 'y']
dy/dx = -3(x^2 - 1) / (x^2 + 1)^2

Answer 2

y=3x/ (x^2+1)

let u = 3x -----> u ' = 3
v = x^2 +1 ------> v ' = 2x

(u/v) ' = (u ' v - u v ')/(v^2)
= (3(x^2 +1) - 3x(2x))/(x^2+1)^2
=(-3x^2 +3) /(x^2 +1)^2
= -3(x^2 -1) / (x^2 +1)^2

Answer:
Derivative of y=3x/ (x^2+1 ) is
= -3(x^2 -1) / (x^2 +1)^2

Answer 3

Brackets should be used to avoid confusion:-
y = (3x) / (x ² + 1)
Quotient rule:-
dy/dx = [ (3) (x ² + 1) - (3x)(2x) ] / (x ² + 1) ²
dy/dx = [ 3 - 3x ² ] / (x² + 1) ²
dy/dx = [ 3 (1 - x ²) ] / (x ² + 1) ²

Answer 4

I think that you are missing some ( ).

What you wrote is:

y = (3x/x^2) + 1 = (3/x) + 1

But I think you meant:

y = 3x/(x^2 + 1 )

Study:
http://en.wikipedia.org/wiki/Table_of_derivatives

Answer 5

Better take log on both sides...n then proceed...