how do I find the derivative of f(x) = (tanx)/(1+x^2)?

Question

how do I find the derivative of f(x) = (tanx)/(1+x^2)

Answer 1

here we use the quotient rule
d/dx(u/v)=(vdu-udv)/v^2

let u=tanx,v=(1+x^2)

d/dx(tanx/(1+x^2))
=(1+x^2)d(tanx)-tanx*d(1+x^2))
/(1+x^2)^2
={(1+x^2)*sec^2(x)-2x*tanx}
/(1+x^2)^2

therefore,if f(x)=tanx/(1+x^2)

f'(x)={(1+x^2)*sec^2(x)
-2x*tanx}/(1+x^2)^2

i hope that this helps

Answer 2

Use the quotient rule which says d(hi/ho) = (hodhi - hidho)/hoho.

More details:

dhi = sec^2(x)
dho = 2x

so (hodhi - hidho)/hoho =

[(1+x^2)sec^2(x) - 2x*tanx]/(1+x^2)^2

I know (hodhi - hidho)/hoho sounds silly, but its the only way I can remember the blasted quotient rule.

Answer 3

let u = tanx and v = 1 + x²
du/dx = sec²x and dv/dx = 2x
By quotient rule:-
f ` (x) = (vdu/dx - udv/dx) / v²
f ` (x) = ((1 + x²)sec²x - tanx .2x)/(1 + x²)²

Answer 4

Using a Ti-89 Titanium (an essential) I get

-(2*x*sinx*cosx-x^2-1)
------------------------------
(x^2+1)^2*(cosx)^2

Answer 5

f(x) = (tan x)*(x²+1)^(-1)
f'(x) = d/dx (tan x) * (x²+1)^(-1) + tan x * d/dx (x²+1)^(-1)
Should be pretty easy after that.