In a question a student was given to find the derivative of the product of two functions $f$ and $g$. The student by mistake thought $(fg)' =f'g'$ for his question $f(x) =x^3$ and he got the correct answer. Given that $g(4)=1$. Which of the following is/are correct possibilities (where $g(x) > 0$)
(A) $g(x)$ is decreasing in $(−∞ ,3)$
(B) $g(x)$ is increasing in $(3, +∞) $
(C) $g(x)$ is decreasing in $(3, +∞)$
(D) $g(x)$ is is increasing in $(−∞,3)$
Obviously, the right way is to use $(fg)' = f'g+fg'$. If the student got the right answer anyway, then for this case of $f$ and $g$:
$$f'g' = f'g+g'f$$
Using $f(x) = x^3$ and letting $g$ be the mystery function, then:
$$3x^2\frac{\mathrm dg}{\mathrm dx} = 3x^2g(x)+x^3\frac{\mathrm dg}{\mathrm dx}$$
For $x \neq 0$, factoring simplifies the equation:
$$\frac{\mathrm dg}{\mathrm dx} = g(x)+\frac{1}{3}x\frac{\mathrm dg}{\mathrm dx}$$
From here, this is just a separable differential equation where you can let $y = g(x)$ and use your condition point $(4, 1)$.
