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

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

2021-01-01
$$\displaystyle{y}={x}\sqrt{{{x}^{{2}}+{48}}}$$
$$\displaystyle{y}={x}\cdot{\left\lbrace{x}^{{2}}+{48}\right\rbrace}^{{\frac{{1}}{{2}}}}$$
Taking logs of both sides.
$$\displaystyle{\ln{{\left({y}\right)}}}={\ln{{\left({x}\right)}}}+{\left(\frac{{1}}{{2}}\right)}{\ln{{\left({x}^{{2}}+{48}\right)}}}$$
differentiating both sides wrt x,
$$\displaystyle{\left(\frac{{1}}{{y}}\right)}{y}'=\frac{{1}}{{x}}+{\left(\frac{{1}}{{2}}\right)}\frac{{{2}{x}}}{{{x}^{{2}}+{48}}}$$
$$\displaystyle{y}'=\frac{{y}}{{x}}+\frac{{{y}{x}}}{{{x}^{{2}}+{48}}}$$
$$\displaystyle{y}'=\sqrt{{{x}^{{2}}+{48}}}+{\left({x}^{{2}}\right)}\frac{\sqrt{{{x}^{{2}}+{48}}}}{{{x}^{{2}}+{48}}}$$
$$\displaystyle{y}'=\frac{{{x}^{{2}}+{48}}}{\sqrt{{{x}^{{2}}+{48}}}}+\frac{{{x}^{{2}}}}{\sqrt{{{x}^{{2}}+{48}}}}$$
$$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{2}{x}^{{2}}+{48}}}{\sqrt{{{x}^{{2}}+{48}}}}$$
$$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={2}\frac{{{x}^{{2}}+{24}}}{\sqrt{{{x}^{{2}}+{48}}}}$$

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