Trignometric Calculas?

Question

When differentiating y=tanx/x

The steps go as follows:?
y'=x * sec^2x - tanx * 1 / x^2

But is it possible to cancel out an "x" from the top and bottom & leave it as:

y'= sec^2x - tanx / x ?????

Answer 1

No, as there is no x attached to the tanx term

Answer 2

y = tan x / x
dy/dx = ( x sec ² x - tan x ) / x ²
Which may be expressed as:-
(1 / x) sec ² x - ( 1 / x ² ) tan x (if desired)
Your differentiation is correct but cancelling is incorrect.

Answer 3

(ax - b) / x^2 doesnt equal (a - b) / x
You can't do that.