Question

If $y=(\log x{)}^x$ then $\frac{dy}{dx}=$

• A. $(\log x{)}^x\left[\log \right(\log x)+\frac{1}{\log x}]$
• B. $(\log x{)}^x\left[\log \right(\log x)-\frac{1}{\log x}]$
• C. $-(\log x{)}^x\left[\log \right(\log x)+\frac{1}{\log x}]$
• D. $-(\log x{)}^x\left[\log \right(\log x)-\frac{1}{\log x}]$

Answer 1

Answer: (A)

$(\log x{)}^x\left[\log \right(\log x)+\frac{1}{\log x}]$

$y=(\log x{)}^x$ then $\frac{dy}{dx}-$

$\log y=x\log \left(\log \right(x\left)\right)$ [Taking log on both sides]

Differentiating on both sides

$\frac{1}{y}\frac{dy}{dx}=\log \left(\log \right(x\left)\right)+\frac{x}{\log \left(x\right)}\frac{d}{dx}\left(\log \right(x\left)\right)$

$\frac{dy}{dx}=y\left[\log \right(\log \left(x\right))+\frac{x}{\log \left(x\right)}\times \left(\frac{1}{x}\right)]$

Substituting y

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