The expression \[\frac{d^2}{dx^2}\bigl[(2x-1)^4(1-x)^5\bigr] + \frac{d}{dx}\bigl[(2x+1)^4(3x^2-2)^2\bigr]\] is a polynomial of degree
\(9\);
\(8\);
\(7\);
less than \(7\).
Let’s observe that
multiplying two polynomials of degrees \(p\) and \(q\) gives a polynomial of degree \(p+q\), and
differentiating a polynomial of degree \(p\) gives a polynomial of degree \(p-1\).
From this we can quickly see that the expression in the question is a sum of two polynomials, each of degree 7.
Does this mean their sum has to be a polynomial of degree 7 too?
Yes, unless the leading terms of the two expressions cancel when you add them together.
So what are our leading terms here?
The expression \((2x-1)^4(1-x)^5\) gives first term \((2x)^4(-x)^5=-16x^9\), which differentiates twice to \[-9 \times 8 \times 16 x^7.\]
The leading term of the expression \((2x+1)^4(3x^2-2)^2\) is \((2x)^4(3x^2)^2=16 \times 9x^8\), which differentiates to \[16 \times 9 \times 8x^7.\]
So in fact the two leading terms here DO cancel, and we’re left with a polynomial of degree less than \(7\), so the answer is (d).
