Jamie Medina

2022-11-26

Let $f$ and $g$ be differentiable functions with following properties: If $h\left(x\right)=f\left(x\right)g\left(x\right)$ and $h\text{'}\left(x\right)=f\left(x\right)g\text{'}\left(x\right)$, then what is $f\left(x\right)$ ?

Aldo Rios

Expert

$h\left(x\right)=f\left(x\right)g\left(x\right)$
differentiate w.r.t ‘$x$
$h\text{'}\left(x\right)=\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)\phantom{\rule{0ex}{0ex}}=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)\left[\because \frac{d}{dx}uv=u\text{'}v+v\text{'}u\right]$
substitute $h\text{'}\left(x\right)=f\left(x\right)g\text{'}\left(x\right)$ in above equation
$f\left(x\right)g\text{'}\left(x\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)\left[\text{Given}\right]\phantom{\rule{0ex}{0ex}}f\text{'}\left(x\right)g\left(x\right)=0$
Since $g\left(x\right)>0$ for all $x$ is given
So $f\text{'}\left(x\right)=0$
It means $f\left(x\right)$ is some constant
$⇒f\left(x\right)=C$
Given $f\left(0\right)=1$
Which means $C=1$
Therefore $f\left(x\right)=1$

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