Jamie Medina

Answered

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 $f\left(x\right)$ ?

Answer & Explanation

Aldo Rios

Expert

2022-11-27Added 8 answers

$h\left(x\right)=f\left(x\right)g\left(x\right)$

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

$\Rightarrow f\left(x\right)=C$

Given $f\left(0\right)=1$

Which means $C=1$

Therefore $f\left(x\right)=1$

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

$\Rightarrow f\left(x\right)=C$

Given $f\left(0\right)=1$

Which means $C=1$

Therefore $f\left(x\right)=1$

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