Is there any function in the form f(x)g(x) in which its derivative is f′(x)g′(x) ?

Answer 1

Here is a simple example where this is true:

f(x) = x

g(x) = 1/(x-1)

f'(x) = 1

g'(x) = -1/(x-1)²

f(x)g(x) = x/(x-1) and taking its derivative using the quotient rule: [(x-1)(1) - (x)(1)]/(x-1)² = -1/(x-1)².

A general solution looks like this:

f'(x)g'(x) = f(x)g'(x) + g(x)f'(x) -- product rule on the right

f(x)g'(x) = f'(x)g'(x) - g(x)f'(x)

f(x)g'(x) = f'(x)[g'(x) - g(x)]

g'(x) = [f'(x)/f(x)] [g'(x) - g(x)]

g'(x)/[g'(x) - g(x)] = f'(x)/f(x)

or

f'(x)/f(x) = g'(x)/[g'(x) - g(x)], that is we separated variables

The antiderivative of the left side is ln|f(x)|. Whether or not you can find an antiderivative on the right side depends on the choice for g(x).

ln|f(x)| + C = ∫ g'(x)dx/[g'(x) - g(x)]

Let g(x) = x, then use the above to solve for f(x).

g'(x) = 1

∫ 1dx/(1 - x) = -ln|1-x| = ln|(1-x)^-1| = ln|1/(x-1)|

So,

ln|f(x)| = ln|1/(x-1)| + C ; let C = ln|k| -> ln|k/(x -1)|

f(x) = k/(x-1)

Let g(x) = e^ax, substitute into the formula to see if you can find an antiderivative and a function for f(x).