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=

Question

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=

freckles · Answer

derivative of (4+x) is 1.

freckles · Answer

you wrote 4

anonymous · Answer

That is what my professor gave me...

freckles · Answer

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

anonymous · Answer

He wants us to work this out by hand.

freckles · Answer

your professor probably made a little mistake then

freckles · Answer

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

freckles · Answer

@ElfQueen are you understanding?