Finding the differential

Question

A homework assignment has had me solve for $dy$ in the equation

$y=\frac{3x}{1+5x^2}$

for which I used the quotient rule, and solved, getting

$\frac{dy}{dx}=\frac{3-15x^2}{(1+5x^2)^2}$

I factored out $-3$ from the top giving me $-3(1+5x^2)$, which there are two of on the bottom, so I removed the term on top, and one of the terms on the bottom, leaving me with

$\frac{-3}{1+5x^2}$ before moving the $dx$ to the end.

The online homework insists I got it wrong,

Did I do something wrong in my simplification of this or did it not have that in it's acceptable answers? I did not have to enter the dx in the answer, it was already given on screen.

Answer 1

Your simplification is not correct. $3-15x^2 = -3(-1+5x^2)$, not what you wrote.

Answer 2

Your answer is almost correct. Don't forget that when you cancel a term, you must state something about the denominator not being zero.

Answer 3

$y’=\frac{3(1+5x^2)-(10x)(3x)}{(1+5x^2)^2}$

$=\frac{3+15x^2 -30x^2}{(1+5x^2)}=\frac{3-15x^2}{(1+5x^2)}$

Notice that you can’t factor out a negative 3, it would make it equal to $-3(-1+5x^2)$ so they won’t cancel with the bottom