Can you derive these...?

Question

f(x) = x^2 / (1+ x^2)

and...

sin 2x - x

Correction:
f(x) = sin 2x - x
but I'm sure you assumed that...

Answer 1

f(x) = x^2 / (1 + x^2)

Quotient Rule:
f(x) = g(x) / h(x)
f'(x) = [g'h - gh'] / h^2

g(x) = x^2
g'(x) = 2x

h(x) = 1 + x^2
h'(x) = 2x

g'h = (2x)(1 + x^2) = 2x + 2x^3
gh' = (x^2)(2x) = 2x^3

So...
g'h - gh' = (2x + 2x^3) - 2x^3 = 2x

f'(x) = (2x) / (1 + x^2)^2

************
f(x) = (sin 2x) - x

f'(x) = (cos 2x)(2) - 1
= 2(cos 2x) - 1

Answer 2

the derivative of the top times the bottom (2x)(1+x^2), minus the derivative of the bottom times the top(2x)(x^2), all over the bottom squared! (1+x^2)(1+x^2)...then it is all just algebra.

the second is using difference rule:(D[u-v] = D[u]-D[v), chain rule: D[f (u)] = f '(u)*D[u], and trig rule for sines: D[sin(x)] = cos(x)

hope that helped!

Answer 3

Question 1
f (x) = x ² / (1 + x ² )
f `(x) = [ 2x (1 + x ² ) - ( x ² )( 2x ) ] / (1 + x ²) ²
f `(x) = ( 2x ) / (1 + x ² ) ²

Question 2
f (x) = sin(2x) - x
f `(x) = 2 cos 2x - 1

Answer 4

f(x) =1-1/(1+x^2)
f´(x) = 2x/(1+x^2)^2
y´= 2cos2x-1