Question

If \(y(x) = (x^x)^x, \quad x > 0\), then \(\frac{d^2x}{dy^2} + 20\) at \(x = 1\) is equal to __________.

Answer 1

Correct Answer: 16

\(∵\) \(y(x) = (x^x)^x\)
\(∴\) \(y=x^{x^2}\)
\(∴\) \(\frac{dy}{dx} = x^2 \cdot x^{x^2 - 1} + x ^{x^2} \ln(x) \cdot 2x\)
\(∴\) …\(\frac{dx}{dy} = \frac{1}{x^{2} + 1(1 + 2\ln x)}\)\(...........(i)\)
Now,
\(\frac{d^2x}{dx^2} = \frac{d}{dx}\left((x^{x^2} + 1(1 + 2\ln x))^{-1}\right) \cdot \frac{dx}{dy}\)

\(=\) \(\frac{-x(x^{x^2}+1(1+2\ln x))^{-2} \cdot x^{ x^2}(1+2\ln x)(x^2+2x^2\ln x+3)}{x^{x^2} \cdot (1+2\ln x)}\)

\(=\) \(-\frac{x^{ x^2}(1+2\ln x)(x^2+3+2x^2\ln x)}{(x^{x^2} \cdot (1+2\ln x))^3}\)

\(\frac{d^2x}{dy^2( at \ x=1)}=−4\)

\(∴\) \(\frac{d^2x}{dy2( at \ ^x=1)}+20=16\)