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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=

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  1. freckles
    • one year ago
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    derivative of (4+x) is 1.

  2. freckles
    • one year ago
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    you wrote 4

  3. anonymous
    • one year ago
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    That is what my professor gave me...

  4. freckles
    • one year ago
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    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
    • one year ago
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    He wants us to work this out by hand.

  6. freckles
    • one year ago
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    your professor probably made a little mistake then

  7. freckles
    • one year ago
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    \[f'(x)=\frac{(4+x)(2x)-x^2 \color{red}{(1)}}{(4+x)^2}\]

  8. freckles
    • one year ago
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    @ElfQueen are you understanding?

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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