Question

Derivative of y=(1+4x)^5(3+x-x^2)^8

Answer 1

Product rule plus chain rule on this one. Let
\[f(x) = (1+4x)^5\]
\[g(x) = (2+x-x^2)^8\]
\[y=f(x)g(x) \rightarrow y' = f'(x)g(x)+f(x)g'(x) \]
now,
\[f'(x)=5(1+4x)^4 * 4 = 20(1+4x)^4\]
and
\[g'(x) = 8(2+x-x^2)^7(-2x+1) \]
so finally,
\[y' = 20(1+4x)^4(2+x-x^2)^8 + 8(1-2x)(2+x-x^2)^7(1+4x)^5\]

Answer 2

I'd like to point out that in your g(x) you have: (3+x-x^2)^8 and not (2+x-x^2)^8 like Jemurray3 wrote. This will simply change that one 2 to a 3...Also, you should simplify your answers.
\[\frac{d}{dx}=20 (4 x+1)^4 (-x^2+x+3)^8+8 (4 x+1)^5 (1-2 x) (-x^2+x+3)^7\]
\[=4 (1 + 4 x)^4 (-3 - x + x^2)^7 (-17 - 9 x + 21 x^2)\]