Question

f(x)=ax^3+bx^2+cx+d (a,b,c,d constant) and
f(x)= (1/a)(x^2+(1/b)x+c) (a,b,c constant)
How do you find f '(x)?

The answer to the first one is: 12x(3x^2+1) but I don't know how to get there. Can someone help please?

Answer 1

\[f(x)=ax^3+bx^2+cx+d\]
Differentiate term by term, using the power rule for each term with an x, and the constant rule for d:
\[f'(x)=3ax^{3-1}+2bx^{2-1}+cx^{1-1}+\frac{d}{dx}d\\
f'(x)=3ax^{2}+2bx^{1}+cx^{0}+0\\
f'(x)=3ax^{2}+2bx+c\]
For the second function:
\[f(x)=\frac{1}{a}\left(x^2+\frac{1}{b}x+c\right)\]
Go ahead and distribute the coefficients, then differentiate term by term like in the first function:
\[f(x)=\frac{1}{a}x^2+\frac{1}{ab}x+\frac{c}{a}\\
f'(x)=\frac{2}{a}x^{2-1}+\frac{1}{ab}x^{1-1}+\frac{d}{dx}\frac{c}{a}\\
f'(x)=\frac{2}{a}x^{1}+\frac{1}{ab}x^{0}+0\\
f'(x)=\frac{2}{a}x+\frac{1}{ab}\]
I'm not sure how the answer for the first is supposed to be \(12x\left(3x^2+1\right)\), since you're not given any information about the constants \(a,b,c,d\).

At first, I thought that maybe the two given functions are the same, but that's not possible, since
\[ax^3+bx^2+cx+d\not= \frac{1}{a}\left(x^2+\frac{1}{b}x+c\right)\]
because the left side is a cubic and the right side is a quadratic.