William Curry

Answered

2021-12-25

For the following exercises, use logarithmic differentiation to find $\frac{dy}{dx}$

$y={({x}^{2}-1)}^{\mathrm{ln}x}$

Answer & Explanation

Donald Cheek

Expert

2021-12-26Added 41 answers

We have given that

$y={({x}^{2}-1)}^{\mathrm{ln}x}$ --- 1

Taking logarithm both side of 1 we get

$\mathrm{ln}y={\mathrm{ln}({x}^{2}-1)}^{\mathrm{ln}x}$

$\Rightarrow \mathrm{ln}y=\left(\mathrm{ln}x\right)\left[\mathrm{ln}({x}^{2}-1)\right]$

Now differentiating both side with respect to x -

$\frac{d}{dx}\left(\mathrm{ln}y\right)=\frac{d}{dx}\left[\mathrm{ln}x\left(\mathrm{ln}({x}^{2}-1)\right)\right]$

$\Rightarrow \frac{1}{y}\cdot \frac{dy}{dx}=\left(\mathrm{ln}x\right)\frac{d}{dx}\left[\mathrm{ln}x({x}^{2}-1)\right]+\mathrm{ln}({x}^{2}-1)\frac{d}{dx}\left(\mathrm{ln}x\right)$

$\Rightarrow \frac{1}{y}\cdot \frac{dy}{dx}=\left(\mathrm{ln}x\right)\frac{1}{{x}^{2}-1}\cdot \frac{d}{dx}({x}^{2}-1)+\mathrm{ln}({x}^{2}-1)\frac{1}{x}$

$\Rightarrow \frac{1}{y}\cdot \frac{dy}{dx}=\frac{\mathrm{ln}x}{{x}^{2}-1}\cdot \left(2x\right)+\frac{\mathrm{ln}({x}^{2}-1)}{x}$

$\Rightarrow \frac{dy}{dx}=y[\frac{2x\mathrm{ln}x}{{x}^{2}-1}+\frac{\mathrm{ln}({x}^{2}-1)}{x}]$

$\Rightarrow \frac{dy}{dx}={({x}^{2}-1)}^{\mathrm{ln}x}[\frac{2x\mathrm{ln}x}{{x}^{2}-1}+\frac{\mathrm{ln}({x}^{2}-1)}{x}]$

Taking logarithm both side of 1 we get

Now differentiating both side with respect to x -

Corgnatiui

Expert

2021-12-27Added 35 answers

Possible derivation:

$\frac{d}{dx}\left({(-1+{x}^{2})}^{\mathrm{log}\left(x\right)}\right)$

Express$({x}^{2}-1)}^{\mathrm{log}\left(x\right)$ as a power of $e:{({x}^{2}-1)}^{\mathrm{log}\left(x\right)}={e}^{\mathrm{log}\left({({x}^{2}-1)}^{\mathrm{log}\left(x\right)}\right)}={e}^{\mathrm{log}\left(x\right)\mathrm{log}({x}^{2}-1)}:$

$=\frac{d}{dx}\left({e}^{\mathrm{log}\left(x\right)\mathrm{log}(-1+{x}^{2})}\right)$

Using the chain rule,$\frac{d}{dx}\left({e}^{\mathrm{log}\left(x\right)\mathrm{log}({x}^{2}-1)}\right)=\frac{d{e}^{u}}{du}\frac{du}{dx}$ , where $u=\mathrm{log}\left(x\right)\mathrm{log}({x}^{2}-1)$ and $\frac{d}{du}\left({e}^{u}\right)={e}^{u}$ :

$=\left(\frac{d}{dx}\left(\mathrm{log}\left(x\right)\mathrm{log}(-1+{x}^{2})\right)\right){e}^{\mathrm{log}\left(x\right)\mathrm{log}(-1+{x}^{2})}$

Express$e}^{\mathrm{log}\left(x\right)\mathrm{log}({x}^{2}-1)$ as a power of $x:{e}^{\mathrm{log}\left(x\right)\mathrm{log}({x}^{2}-1)}={e}^{\mathrm{log}\left({x}^{\mathrm{log}({x}^{2}-1)}\right)}={x}^{\mathrm{log}({x}^{2}-1)}:$

$=\left({x}^{\mathrm{log}(-1+{x}^{2})}\right)\frac{d}{dx}\left(\mathrm{log}\left(x\right)\mathrm{log}(-1+{x}^{2})\right)$

Use the product rule,$\frac{d}{dx}\left(uv\right)=v\frac{du}{dx}+u\frac{ddiv>}{}$

Express

Using the chain rule,

Express

Use the product rule,

0

user_27qwe

Expert

2021-12-30Added 230 answers

First answer is right!

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