Two Sure Questions from Logarithmic Differentiation

This lesson works through two differentiation questions that both use logarithmic differentiation: an implicit equation in $x$ and $y$, and the value of a derivative of a product of four factors at a point.

Problem 1: Differentiate $x^y + y^x = 1$

We are given

$x^y + y^x = 1$

and we want $\tfrac{dy}{dx}$. Set $u = x^y$ and $v = y^x$, so the equation becomes

$u + v = 1.$

Differentiating with respect to $x$,

$\frac{du}{dx} + \frac{dv}{dx} = 0. \qquad (1)$

We now find $\tfrac{du}{dx}$ and $\tfrac{dv}{dx}$ separately.

Differentiating $u = x^y$

Take logs on both sides:

$\log u = y \log x.$

Differentiate with respect to $x$:

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

Multiply through by $u = x^y$:

$\frac{du}{dx} = x^y\left(\frac{y}{x} + \log x \,\frac{dy}{dx}\right). \qquad (2)$

Differentiating $v = y^x$

Take logs on both sides:

$\log v = x \log y.$

Differentiate with respect to $x$, applying the product rule on the right:

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

Multiply through by $v = y^x$:

$\frac{dv}{dx} = y^x\left(\frac{x}{y}\frac{dy}{dx} + \log y\right). \qquad (3)$

Combining and solving for $\tfrac{dy}{dx}$

Substitute $(2)$ and $(3)$ into $(1)$:

$x^y\left(\frac{y}{x} + \log x \,\frac{dy}{dx}\right) + y^x\left(\frac{x}{y}\frac{dy}{dx} + \log y\right) = 0.$

Collect the $\tfrac{dy}{dx}$ terms on one side and move everything else to the right:

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

Therefore

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

Problem 2: Find $f'(1)$ for a product of four factors

We are given

$f(x) = (1+x)(1+x^2)(1+x^3)(1+x^4)$

and we want $f'(1)$. Because this is a product of more than two factors, we use logarithmic differentiation.

Take logs to turn the product into a sum

$\log f(x) = \log(1+x) + \log(1+x^2) + \log(1+x^3) + \log(1+x^4).$

Differentiate term by term

$\frac{1}{f(x)}f'(x) = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{3x^2}{1+x^3} + \frac{4x^3}{1+x^4}.$

Multiply by $f(x)$:

$f'(x) = (1+x)(1+x^2)(1+x^3)(1+x^4)\left(\frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{3x^2}{1+x^3} + \frac{4x^3}{1+x^4}\right).$

Substitute $x = 1$

Each factor becomes $1 + 1 = 2$, so the product is $2\cdot 2\cdot 2\cdot 2 = 16$. The bracket becomes

$\frac{1}{2} + \frac{2}{2} + \frac{3}{2} + \frac{4}{2} = \frac{1+2+3+4}{2} = \frac{10}{2} = 5.$

Hence

$f'(1) = 16 \times 5 = 80.$

Key takeaways

• For an implicit equation like $x^y + y^x = 1$, split it into named parts, take the log of each part, and differentiate before combining.
• When a power has both its base and exponent depending on $x$, taking logs first lets you apply the product rule cleanly.
• Taking the log of a product turns it into a sum, so each factor can be differentiated on its own; here $f'(1) = 80$.