Use logarithmic differentiation to find dy/dx y=x sqrt{x^2+48}

Question

Y= x times the sqrt of x^2+48

Answer 1

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}