# The derivative has a higher grade than the function itself. How is that possible?

I can't solve a question of a test of a pre-university mathematics course. I understand the rules of derivatives but I am blocked when trying to solve the following question below.

I tried to solve the question by making the derivative of the function and equaling that to the derivative in the question. So, I got that $a= \frac{21}{x^4}$ etc. But, I may not use those answers as solution for the question.

What am I seeing wrong or where am I blind? Could someone give me a start to solve this question?

This is the question:

If $$f(x)=7x^3-2x^2+4x-11$$ then de derivative $$D(f(x))=ax^4+bx^3+cx^2+dx+e$$ What are the values for $a,b,c,d$ and $e$?

• Is there any restriction on $a,b,c,d,e$? Why not $a=b=0$? Also, I think $a=\frac{21}{x^4}$ would not be correct because one typically assumes they are constants, while you are allowing it to be a function of $x$. – GoodDeeds Sep 4 '18 at 5:42
• From the differentiation you should get $$f'(x) = 21x^2 - 4 x +4 = 0x^4 + 0x^3 + 21x^2 -4x +4$$ – Matti P. Sep 4 '18 at 5:45
The coefficients $a,b,c,d,e$ must be constants. They cannot be dependent on $x$. I think you've gotten hung up on the fact there seems to be a quartic term $(ax^4)$, which you feel shouldn't be there. You can make these terms disappear with zero-coefficients. By ordinary differentiation, you get that $$D(7x^3-2x^2+4x-11) = 21x^2-4x+4 = \underbrace{0}_{a}x^4 + \underbrace{0}_{b}x^3 + \underbrace{21}_{c}x^2 + \underbrace{(-4)}_{d}x + \underbrace{4}_{e}$$
Therefore, the values are $a = 0$, $b = 0$, $c = 21$, $d = -4$, and $e = 4$.