Question

If y = (log x)^x + x^log x, find ${\displaystyle \frac{\text{dy}}{\text{dx}}.}$

Answer 1

Let y =(log x)^x + x^log x

Also, let u =(log x)^x and  v = x^log x

∴ y = u + v

${\displaystyle \Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}}$ ........(1)

u = (logx)^x

⇒ log u = log[(log x)^x]

⇒ log u = x log(log x)

Differentiating both sides with respect to x, we obtain

${\displaystyle \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\left(\text{x}\right)\times \log \left(\log \text{x}\right)+\text{x}.\frac{\text{d}}{\text{dx}}\left[\log \left(\log \text{x}\right)\right]}$

${\displaystyle \Rightarrow \frac{\text{du}}{\text{dx}}=\text{u}[1\times \log \left(\log \text{x}\right)+\text{x}.\frac{1}{\log \text{x}}.\frac{\text{d}}{\text{dx}}\left(\log \text{x}\right)]}$

${\displaystyle \Rightarrow \frac{\text{du}}{\text{dx}}={\left(\log \text{x}\right)}^{\text{x}}[\log \left(\log \text{x}\right)+\frac{\text{x}}{\log \text{x}}.\frac{1}{\text{x}}]}$

${\displaystyle \Rightarrow \frac{\text{du}}{\text{dx}}={\left(\log \text{x}\right)}^{\text{x}}[\log \left(\log \text{x}\right)+\frac{1}{\log \text{x}}]}$

${\displaystyle \Rightarrow \frac{\text{du}}{\text{dx}}={\left(\log \text{x}\right)}^{\text{x}}\left[\frac{\log \left(\log \text{x}\right).\log \text{x}+1}{\log \text{x}}\right]}$

${\displaystyle \Rightarrow \frac{\text{du}}{\text{dx}}={\left(\log \text{x}\right)}^{\text{x}-1}[1+\log \text{x}.\log \left(\log \text{x}\right)]}$ .....(2)

v = x^logx

⇒ log v = log(x^logx)

⇒ log v = log  x log x = (log x)²

Differentiating both sides with respect to x, we obtain

${\displaystyle \frac{1}{\text{v}}.\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\left[{\left(\log \text{x}\right)}^2\right]}$

${\displaystyle \Rightarrow \frac{1}{\text{v}}.\frac{\text{dv}}{\text{dx}}=2\left(\log \text{x}\right).\frac{\text{d}}{\text{dx}}\left(\log \text{x}\right)}$

${\displaystyle \Rightarrow \frac{\text{dv}}{\text{dx}}=2\text{v}\left(\log \text{x}\right).\frac{1}{\text{x}}}$

${\displaystyle \Rightarrow \frac{\text{dv}}{\text{dx}}=2{\text{x}}^{\log \text{x}}\frac{\log \text{x}}{\text{x}}}$

${\displaystyle \Rightarrow \frac{\text{dv}}{\text{dx}}=2{\text{x}}^{\log \text{x}-1}.\log \text{x}}$ ......(3)

Therefore, from (1), (2), and (3), we obtain

${\displaystyle \frac{\text{dy}}{\text{dx}}={\left(\log \text{x}\right)}^{\text{x}-1}[1+\log \text{x}.\log \left(\log \text{x}\right)]+2{\text{x}}^{\log \text{x}-1}.\log \text{x}}$