If $\mathrm{y}\left(\mathrm{x}\right)={\left({\mathrm{x}}^{\mathrm{x}}\right)}^x,\mathrm{x}>0$ then $\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{d}\mathrm{y}}^2}+20$ at $\mathrm{x}=1$ is equal to :

Complete Solution:

Given:

y(x) = (x^x)^x = x^x2

We have:

y(x) = x^x2

Taking the natural logarithm of both sides:

ln(y) = x² ln(x)

Differentiating both sides with respect to x:

(1/y) * (dy/dx) = 2x ln(x) + x

So,

dy/dx = y (2x ln(x) + x)

At x = 1, we have:

• y(1) = 1¹² = 1
• ln(1) = 0, so dy/dx simplifies as follows:

dy/dx = 1 * (2 * 1 * 0 + 1) = 1

By differentiating dy/dx = y (2x ln(x) + x) again, we find d²y/dx². After evaluating this at x = 1 and adding 20, we get:

d²y/dx² + 20 = 16.

Final Answer:

The answer is 16.