Hello, I am having trouble finding a valid function for g(x) to make the following statement true. f'g'=(fg)', let f(x)=e^(x^2)

Question

Hello, I am having trouble finding a valid function for the about question. I have been using the product rule and dividing by e but cannot seem to get a correct function to fit the rule.

Answer 1

Assuming that f(x)=exp(x²) the equation f''g'=(fg)'=f'g+fg'  now reads

2xexp(x²)g'=2xexp(x²)g+exp(x²)g'

(2x-1)exp(x²)g'=2xexp(x²)g since exp(x²) is never 0 

(2x-1)g'=2xg

g'/g=2x/(2x-1)=(2x-1+1)/(2x-1)=1+1/(2x-1)  Integrating we have

logg(x)=x+(1/2)log(2x-1)+constant

 

g(x)=exp(x+(1/2)log(2x-1)+constant)

 

Answer 2

In general, (fg)' = f'g + fg' where both f and g are functions of x. This is called the product rule.

 

If f(x) = e^(x²), then f' = 2x•e^(x²), by use of the chain rule.

 

Clarify your thinking! (abandon your search).