William Curry

2021-12-25

For the following exercises, use logarithmic differentiation to find $\frac{dy}{dx}$
$y={\left({x}^{2}-1\right)}^{\mathrm{ln}x}$

Donald Cheek

Expert

We have given that
$y={\left({x}^{2}-1\right)}^{\mathrm{ln}x}$ --- 1
Taking logarithm both side of 1 we get
$\mathrm{ln}y={\mathrm{ln}\left({x}^{2}-1\right)}^{\mathrm{ln}x}$
$⇒\mathrm{ln}y=\left(\mathrm{ln}x\right)\left[\mathrm{ln}\left({x}^{2}-1\right)\right]$
Now differentiating both side with respect to x -
$\frac{d}{dx}\left(\mathrm{ln}y\right)=\frac{d}{dx}\left[\mathrm{ln}x\left(\mathrm{ln}\left({x}^{2}-1\right)\right)\right]$
$⇒\frac{1}{y}\cdot \frac{dy}{dx}=\left(\mathrm{ln}x\right)\frac{d}{dx}\left[\mathrm{ln}x\left({x}^{2}-1\right)\right]+\mathrm{ln}\left({x}^{2}-1\right)\frac{d}{dx}\left(\mathrm{ln}x\right)$
$⇒\frac{1}{y}\cdot \frac{dy}{dx}=\left(\mathrm{ln}x\right)\frac{1}{{x}^{2}-1}\cdot \frac{d}{dx}\left({x}^{2}-1\right)+\mathrm{ln}\left({x}^{2}-1\right)\frac{1}{x}$
$⇒\frac{1}{y}\cdot \frac{dy}{dx}=\frac{\mathrm{ln}x}{{x}^{2}-1}\cdot \left(2x\right)+\frac{\mathrm{ln}\left({x}^{2}-1\right)}{x}$
$⇒\frac{dy}{dx}=y\left[\frac{2x\mathrm{ln}x}{{x}^{2}-1}+\frac{\mathrm{ln}\left({x}^{2}-1\right)}{x}\right]$
$⇒\frac{dy}{dx}={\left({x}^{2}-1\right)}^{\mathrm{ln}x}\left[\frac{2x\mathrm{ln}x}{{x}^{2}-1}+\frac{\mathrm{ln}\left({x}^{2}-1\right)}{x}\right]$

Corgnatiui

Expert

Possible derivation:
$\frac{d}{dx}\left({\left(-1+{x}^{2}\right)}^{\mathrm{log}\left(x\right)}\right)$
Express ${\left({x}^{2}-1\right)}^{\mathrm{log}\left(x\right)}$ as a power of $e:{\left({x}^{2}-1\right)}^{\mathrm{log}\left(x\right)}={e}^{\mathrm{log}\left({\left({x}^{2}-1\right)}^{\mathrm{log}\left(x\right)}\right)}={e}^{\mathrm{log}\left(x\right)\mathrm{log}\left({x}^{2}-1\right)}:$
$=\frac{d}{dx}\left({e}^{\mathrm{log}\left(x\right)\mathrm{log}\left(-1+{x}^{2}\right)}\right)$
Using the chain rule, $\frac{d}{dx}\left({e}^{\mathrm{log}\left(x\right)\mathrm{log}\left({x}^{2}-1\right)}\right)=\frac{d{e}^{u}}{du}\frac{du}{dx}$, where $u=\mathrm{log}\left(x\right)\mathrm{log}\left({x}^{2}-1\right)$ and $\frac{d}{du}\left({e}^{u}\right)={e}^{u}$:
$=\left(\frac{d}{dx}\left(\mathrm{log}\left(x\right)\mathrm{log}\left(-1+{x}^{2}\right)\right)\right){e}^{\mathrm{log}\left(x\right)\mathrm{log}\left(-1+{x}^{2}\right)}$
Express ${e}^{\mathrm{log}\left(x\right)\mathrm{log}\left({x}^{2}-1\right)}$ as a power of $x:{e}^{\mathrm{log}\left(x\right)\mathrm{log}\left({x}^{2}-1\right)}={e}^{\mathrm{log}\left({x}^{\mathrm{log}\left({x}^{2}-1\right)}\right)}={x}^{\mathrm{log}\left({x}^{2}-1\right)}:$
$=\left({x}^{\mathrm{log}\left(-1+{x}^{2}\right)}\right)\frac{d}{dx}\left(\mathrm{log}\left(x\right)\mathrm{log}\left(-1+{x}^{2}\right)\right)$
Use the product rule,

user_27qwe

Expert