Maths help, can anyone help me to solve this?

Question

A student forgot the Product Rule for differentiation and made the mistake of
thinking that (fg)' = f'g'. However, they were lucky and got the correct answer.
The function f that they used was f(x) = e^(x^2) and the domain of the problem was
the interval ( 1/2 , infinity). What was the function g?

Answer 1

So you want that:

f'(x)g'(x) = f'(x)g(x) + f(x)g'(x)

Divide both sides by f(x)g(x), and you get:

(f'(x)/f(x)) (g'(x)/g(x)) = f'(x)/f(x) + g'(x)/g(x)

In general, if H(x) = ln h(x), then H'(x) = h'(x)/h(x).

We can see that f'(x) = 2x * e^(x^2), so f'(x)/f(x) = 2x,
and therefore we need to find a g(x) so that:

2x* g'(x)/g(x) = 2x + g'(x)/g(x)

Or:

g'(x)/g(x) = 2x/(2x-1) = 1 + 1/(2x-1)

If G(x) = ln(g(x)), then:

G'(x) = 1 + 1/(2x-1)

So:

G(x) = x + (1/2)*ln(2x-1) + C

so:

g(x) = e^G(x) = e^C * e^x * sqrt(2x-1)

Setting C' to e^C, that is:

g(x) = Ce^x * sqrt(2x-1)

for any constant C.

Answer 2

Very nice question:

(e^x^2) ' = 2x e^(x^2)

hence

2x e^(x^2) g + e^(x^2) g' = 2x e^(x^2) g' [product rule gave same result as product of derivatives]

divide by e^(x^2)
2xg + g' = 2x g'
(2x - 1) g' = 2xg
this is a separable differential equation
g'/ g = 2x / (2x - 1)
ln I g I = x + 0.5 ln (2x - 1) + C...I ignored absolute value since domain is greater than 0.5.
g = A sqrt (2x - 1) e^ x.

Answer 3

f'(x)= 2xe^x^2
So (fg)' = f(x)g'(x) +g(x)f'(x) = e^(x^2)(g'(x)) + 2xe^(x^2)(g(x)).
f'g' = 2xe^(x^2)(g'(x)).
So 2xe^(x^2)(g'(x))= e^(x^2)(g'(x)) +2xe^(x^2)(g(x))
2xg'(x) = g'(x) +2x(g(x))
g(x) = (2x-1)(g'(x))/2