Blog: `ln(xy) + 5x = 30` Use implicit differentiation to find dy/dx

Find $\displaystyle\frac{\text{dy}}{\text{dx}}$ if $\displaystyle{\ln{{\left({x}{y}\right)}}}+{5}{x}={30}$ :

Use a property of logarithms to rewrite the first term:

$\displaystyle{\ln{{x}}}+{\ln{{y}}}+{5}{x}={30}$

Now differentiate term by term with respect to x:

$\displaystyle\frac{{1}}{{x}}+\frac{{1}}{{y}}\cdot\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{5}={0}$

$\displaystyle\frac{{1}}{{y}}\cdot\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-{\left(\frac{{1}}{{x}}+{5}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-{y}{\left(\frac{{1}}{{x}}+{5}\right)}$

or $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-\frac{{{y}{\left({1}+{5}{x}\right)}}}{{x}}$

Find $\displaystyle\frac{\text{dy}}{\text{dx}}$ if $\displaystyle{\ln{{\left({x}{y}\right)}}}+{5}{x}={30}$ :

Use a property of logarithms to rewrite the first term:

$\displaystyle{\ln{{x}}}+{\ln{{y}}}+{5}{x}={30}$

Now differentiate term by term with respect to x:

$\displaystyle\frac{{1}}{{x}}+\frac{{1}}{{y}}\cdot\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{5}={0}$

$\displaystyle\frac{{1}}{{y}}\cdot\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-{\left(\frac{{1}}{{x}}+{5}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-{y}{\left(\frac{{1}}{{x}}+{5}\right)}$

or $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-\frac{{{y}{\left({1}+{5}{x}\right)}}}{{x}}$

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