what is the formula for this function???

Question

[ f/(f+g) ]' =???????

Answer 1

Don't panic! 8-) Use the rules you kow: chain rule and quotient rule.

If it helps, let h = f + g. Then
[f / (f+g)]' = [f/h]' = (f'h - h'f) / h^2
= (f'(f+g) - (f' + g')f) / (f+g)^2
= (f'f + f'g - f'f - g'f) / (f+g)^2
= (f'g - g'f) / (f+g)^2.

You can also do this directly:
[f / (f+g)]' = [f'(f+g) - (f+g)'f] / (f+g)^2
= (f'f + f'g - f'f - g'f) / (f+g)^2
= (f'g - g'f) / (f+g)^2.

Answer 2

We use the quotient rule. A reminder that
[f(x)/g(x)]' = [f'(x)g(x) - f(x) g'(x)] / [g(x)]^2

Therefore,

[f(x) / (f(x) + g(x)) ]' =
{ f'(x) [f(x) + g(x)] - f(x) [f'(x) + g'(x)] } / [f(x) + g(x)]^2

I suppose we can simplify this.

[f'(x)f(x) + f'(x) g(x) - f(x)f'(x) - f(x) g'(x)] / [f(x) + g(x)]^2

Stuff cancels on the top, leaving

[f'(x) g(x) - f(x)g'(x)] / [f(x) + g(x)]^2

Answer 3

this isn't so hard
first derive f' inverse function
then derive f+g
divide!
and i suppose u do know how to derive inverse