More questions on logarithmic differentiation

This lesson works through three problems that lean on logarithmic differentiation. The first proves the product rule for a product of three functions, both the direct way and by taking logs. The next two differentiate equations where the variable appears in an exponent, which is exactly where taking logs of both sides pays off.

Product rule for a product of three functions

Let $y = uvw$, where $u$, $v$ and $w$ are functions of $x$. The goal is to show

$\frac{dy}{dx} = u\,v\,\frac{dw}{dx} + v\,w\,\frac{du}{dx} + u\,w\,\frac{dv}{dx}.$

Method 1: ordinary product rule

Treat $uv$ as the first function and $w$ as the second:

$\frac{dy}{dx} = uv\,\frac{dw}{dx} + w\,\frac{d}{dx}(uv).$

Applying the product rule again to $\frac{d}{dx}(uv) = u\,\frac{dv}{dx} + v\,\frac{du}{dx}$ gives

$\frac{dy}{dx} = uv\,\frac{dw}{dx} + wu\,\frac{dv}{dx} + wv\,\frac{du}{dx}.$

Method 2: logarithmic differentiation

Take logs of both sides of $y = uvw$:

$\log y = \log u + \log v + \log w.$

Differentiate with respect to $x$:

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{u}\frac{du}{dx} + \frac{1}{v}\frac{dv}{dx} + \frac{1}{w}\frac{dw}{dx}.$

Multiply through by $y = uvw$. Each term keeps the two factors that are not being differentiated:

$\frac{dy}{dx} = vw\,\frac{du}{dx} + uw\,\frac{dv}{dx} + uv\,\frac{dw}{dx},$

which is the same result as Method 1.

Differentiating $xy = e^{x-y}$

Take logs of both sides:

$\log(xy) = (x - y)\log e = x - y,$

since $\log e = 1$. So $\log x + \log y = x - y$. Differentiate with respect to $x$:

$\frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx}.$

Collect the $\frac{dy}{dx}$ terms:

$\frac{dy}{dx}\left(\frac{1}{y} + 1\right) = 1 - \frac{1}{x} = \frac{x-1}{x}.$

Since $\frac{1}{y} + 1 = \frac{1+y}{y}$, we get

$\frac{dy}{dx} = \frac{y(x-1)}{x(1+y)}.$

Differentiating $y^{x} = x^{y}$

Take logs of both sides:

$x\log y = y\log x.$

Differentiate with respect to $x$, using the product rule on each side:

$x\cdot\frac{1}{y}\frac{dy}{dx} + \log y = \frac{y}{x} + \log x\,\frac{dy}{dx}.$

Collect the $\frac{dy}{dx}$ terms on one side:

$\frac{dy}{dx}\left(\frac{x}{y} - \log x\right) = \frac{y}{x} - \log y.$

So

$\frac{dy}{dx} = \frac{\dfrac{y}{x} - \log y}{\dfrac{x}{y} - \log x}.$

Taking common denominators in the numerator and the denominator and simplifying gives

$\frac{dy}{dx} = \frac{y\,(y - x\log y)}{x\,(x - y\log x)}.$

Key takeaways

• For a product of three functions, $\dfrac{d}{dx}(uvw) = vw\,\dfrac{du}{dx} + uw\,\dfrac{dv}{dx} + uv\,\dfrac{dw}{dx}$, and logs give the same answer as the product rule.
• Taking logs of both sides turns products into sums and brings exponents down as coefficients, which makes implicit equations much easier to differentiate.
• When the variable appears in an exponent, as in $xy = e^{x-y}$ or $y^x = x^y$, logarithmic differentiation is usually the cleanest route.