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