## anonymous one year ago If f(x)= x^2/(4+x)/ find f "(4). This is what I have done so far: f(x)=x^2/(4+x) = f '(x)=(4+x)(2x)-x^2(4)/(4+x)^2=8x+2x^2-4x^2/(4+x)^2=-x^2^2+8x/(4+x)^2= f "(x)=(x^2+16x+16)(-4x+8)-(-2x^2+8x)(-4x+8)/(x^2+16x+16)^2 After that, this is the part that I am stuck on: (-4x+8)(x^2+16x+16--2x^2-8x)/[(x+4)^2]^2=(-4x^2+8)(3x^2+8x+16)/[(x+4)^2]^2=

1. freckles

derivative of (4+x) is 1.

2. freckles

you wrote 4

3. anonymous

That is what my professor gave me...

4. freckles

still doesn't change that fact that d(4+x)/dx=d(4)/dx+d(x)/dx=0+1=1 and not 4...

5. anonymous

He wants us to work this out by hand.

6. freckles

7. freckles

$f'(x)=\frac{(4+x)(2x)-x^2 \color{red}{(1)}}{(4+x)^2}$

8. freckles

@ElfQueen are you understanding?