Question: Use logarithmic differentiation to find dy/dx y=x sqrt{x^2+48}
y = x sqrt{x^2 + 48}
y=x*{x^2 + 48}^(1/2)
Taking logs of both sides.
ln(y) = ln(x) + (1/2)ln(x^2 + 48)
differentiating both sides wrt x,
(1/y)y' = 1/x + (1/2)(2x)/(x^2 + 48)
y' = y/x + (yx)/(x^2 + 48)
y' = sqrt{x^2 + 48} + (x^2)sqrt{x^2 + 48} / (x^2 + 48)
y' = (x^2 + 48) / sqrt{x^2 + 48} + (x^2) / sqrt{x^2 + 48}
dy/dx = (2x^2 + 48) / sqrt{x^2 + 48}
dy/dx = 2(x^2 + 24) / sqrt{x^2 + 48}