Question

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

• A. $\left(x^x{)}^x\right(1+2\log x)$
• B. $\left(x^x{)}^x\right(1-2\log x)$
• C. $x\left(x^x{)}^x\right(1+2\log x)$
• D. $x\left(x^x{)}^x\right(1-2\log x)$

Answer 1

Answer: (C)

$x\left(x^x{)}^x\right(1+2\log x)$

We have,

$y={\left(x^x\right)}^x$

$y=x^{x^2}$

On taking $\log$ both sides, we get

$\log y=x^2\log x$ …… (1)

On differentiating w.r.t $x$, we get

$\frac{1}{y}\frac{dy}{dx}=\frac{x^2}{x}+\log x\left(2x\right)$

$\frac{1}{y}\frac{dy}{dx}=x+2x\log x$

$\frac{dy}{dx}=y(x+2x\log x)$

$\frac{dy}{dx}={\left(x^x\right)}^x(x+2x\log x)$

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

Hence, this is the answer.