[tex]f(x)+x^4f(x)^3=2005[/tex]
Differentiating both sides yields
[tex]\dfrac{\mathrm d}{\mathrm dx}f(x)+\dfrac{\mathrm d}{\mathrm dx}(x^4f(x)^3)=\dfrac{\mathrm d}{\mathrm dx}2005[/tex]
The right hand side is a constant, so the derivative is just 0, and [tex]\dfrac{\mathrm d}{\mathrm dx}f(x)=f'(x)[/tex]. The remaining term requires using the product rule:
[tex]\dfrac{\mathrm d}{\mathrm dx}(x^4f(x)^3)=f(x)^3\dfrac{\mathrm d}{\mathrm dx}x^4+x^4\dfrac{\mathrm d}{\mathrm dx}f(x)^3[/tex]
Using the power rule for the first term here and the power/chain rules for the second gives
[tex]\dfrac{\mathrm d}{\mathrm dx}(x^4f(x)^3)=4x^3f(x)^3+3x^4f(x)^2\dfrac{\mathrm d}{\mathrm dx}f(x)[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dx}(x^4f(x)^3)=4x^3f(x)^3+3x^4f(x)^2f'(x)[/tex]
Putting everything together, you have
[tex]f'(x)+4x^3f(x)^3+3x^4f(x)^2f'(x)=0[/tex]
Collecting the terms with [tex]f'(x)[/tex] gives
[tex](1+3x^4f(x)^2)f'(x)+4x^3f(x)^3=0[/tex]
[tex](1+3x^4f(x)^2)f'(x)=-4x^3f(x)^3[/tex]
[tex]f'(x)=-\dfrac{4x^3f(x)^3}{1+3x^4f(x)^2}[/tex]
Now, finding the value of the derivative at [tex]x=2[/tex] is just a matter of plugging in [tex]x=2[/tex]:
[tex]f'(2)=-\dfrac{4\times2^3f(2)^3}{1+3\times2^4f(2)^2}[/tex]
We know [tex]f(2)=5[/tex], so we get
[tex]f'(2)=-\dfrac{4\times2^3\times5^3}{1+3\times2^4\times5^2}[/tex]
[tex]f'(2)=-\dfrac{4000}{1201}[/tex]
