We have to find y' for y = (2-x)^(sqrt x)

Use natural logariths for both the sides

ln y = ln[ (2-x)^(sqrt x)]

use the property ln a^x = a*ln x

=> ln y = (sqrt x)*ln ( 2 - x)

Do implicit differentiation of both the sides

=> y'/y = -sqrt x/(2 - x) + (1/2)*ln(2 - x)/sqrt x

=> y' = y[(1/2)*ln(2 - x) - x/(2 - x)]/sqrt x

**y' = [(2-x)^(sqrt x)]*[(1/2)*ln(2 - x) - x/(2 - x)]/sqrt x**

This problem requires logarithmic differentiation:

y = (2-x)^sqrt x

We'll take natural logarithms both sides:

ln y = ln [(2-x)^sqrt x]

ln y = sqrt x * ln (2-x)

We'll differentiate both sides, applying product rule to the right side:

(1/y)*(dy/dx) = (1/2sqrtx)*ln (2-x) + sqrt x*[-1/(2-x)]

dy/dx = y*{(1/2sqrtx)*ln (2-x) + sqrt x*[-1/(2-x)]}

But y = (2-x)^sqrt x

**dy/dx = (2-x)^sqrt x*{(1/2sqrtx)*ln (2-x) + sqrt x*[-1/(2-x)]}**