Logarithmic Differentiation

The power rule handles $x^5$ (variable base, constant power) and the exponential rule handles $5^x$ (constant base, variable power). But what about $x^{\sin x}$ — variable to a variable power? Neither rule applies. The escape is to take logarithms first.

For $y = [u(x)]^{v(x)}$:

1. Take logs: $\log y = v(x)\log u(x)$ — the awkward power becomes an ordinary product.
2. Differentiate both sides (chain rule on the left, product rule on the right): $$\frac{1}{y}\frac{dy}{dx} = v(x)\cdot\frac{u'(x)}{u(x)} + v'(x)\log u(x)$$
3. Multiply back by $y$: $$\boxed{\;\frac{dy}{dx} = y\left[\frac{v(x)}{u(x)}u'(x) + v'(x)\log u(x)\right]\;}$$

The fine print: $f(x)$ and $u(x)$ must be positive — otherwise their logarithms are undefined and the whole method has no meaning. Exam problems supply the interval (e.g. $x > 0$ for $x^{\sin x}$) precisely for this reason.

Two instant corollaries

Constant base: $y = a^x$ ($a > 0$): $\log y = x\log a$, so $$\frac{d}{dx}\left(a^x\right) = a^x \log a$$

The classic $x^x$: $\log y = x \log x$, so $\frac{1}{y}y' = \log x + 1$ and $$\frac{d}{dx}\left(x^x\right) = x^x(1 + \log x)$$

Products and quotients love logarithms too

Even with no variable powers in sight, logs turn heavy products into light sums. For $y = \frac{(x-3)(x^2+4)}{3x^2+4x+5}$ under a square root, taking logs gives $$\log y = \frac{1}{2}\left[\log(x-3) + \log(x^2+4) - \log(3x^2+4x+5)\right]$$ and differentiating term by term beats any quotient-rule slog. The keep-it-straight table:

Solved Examples

Example 1 — The constant-base corollary

Differentiate $a^x$ w.r.t. $x$, where $a$ is a positive constant.

Step 1 — take logs: $\log y = x\log a$.

Step 2 — differentiate: $\frac{1}{y}\frac{dy}{dx} = \log a$.

Answer: $\frac{dy}{dx} = a^x\log a$. (Alternatively, $a^x = e^{x\log a}$ and one chain-rule link gives the same.) In particular $\frac{d}{dx}2^x = 2^x\log 2$ — and for $a = e$, $\log e = 1$ recovers $\frac{d}{dx}e^x = e^x$.

Example 2 — The flagship $x^{\sin x}$

Differentiate $x^{\sin x}$, $x > 0$, w.r.t. $x$.

Step 1 — take logs: $\log y = \sin x \log x$.

Step 2 — product rule on the right: $\frac{1}{y}\frac{dy}{dx} = \cos x \log x + \frac{\sin x}{x}$.

Step 3 — multiply back: $$\frac{dy}{dx} = x^{\sin x}\left(\cos x \log x + \frac{\sin x}{x}\right)$$

Answer: as displayed — the answer always carries the original $y = x^{\sin x}$ out front.

Example 3 — A square root of a rational product

Differentiate $\sqrt{\dfrac{(x-3)(x^2+4)}{3x^2+4x+5}}$ w.r.t. $x$.

Step 1 — take logs (the root becomes $\frac{1}{2}$, the product a sum): $$\log y = \frac{1}{2}\left[\log(x-3) + \log(x^2+4) - \log(3x^2+4x+5)\right]$$

Step 2 — differentiate term by term: $$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2}\left[\frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5}\right]$$

Step 3 — multiply back by $y$: $$\frac{dy}{dx} = \frac{1}{2}\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}\left[\frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5}\right]$$

Answer: as displayed — three logarithm laws (root, product, quotient) converted a quotient-rule nightmare into three easy fractions.

Example 4 — Powers of a logarithm

Differentiate $(\log x)^{\cos x}$, $x > 1$, w.r.t. $x$.

Step 1 — take logs: $\log y = \cos x \log(\log x)$.

Step 2 — differentiate (product rule + chain rule): $$\frac{1}{y}\frac{dy}{dx} = -\sin x \log(\log x) + \cos x \cdot \frac{1}{\log x}\cdot\frac{1}{x}$$

Answer: $\dfrac{dy}{dx} = (\log x)^{\cos x}\left[\dfrac{\cos x}{x\log x} - \sin x\log(\log x)\right]$ — same template, exotic-looking base.

Example 5 — A three-factor product, no variable powers

Differentiate $y = (x+3)^2(x+4)^3(x+5)^4$ w.r.t. $x$.

Step 1 — take logs: $\log y = 2\log(x+3) + 3\log(x+4) + 4\log(x+5)$.

Step 2 — differentiate: $\frac{1}{y}\frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}$.

Answer: $\dfrac{dy}{dx} = (x+3)^2(x+4)^3(x+5)^4\left[\dfrac{2}{x+3} + \dfrac{3}{x+4} + \dfrac{4}{x+5}\right]$ — a triple product rule condensed into one line of fractions.

Example 6 — Mixed implicit and logarithmic

Find $\dfrac{dy}{dx}$ if $y^x + x^y + x^x = a^b$, for $x, y > 0$.

Step 1 — differentiate each term with its own log-trick: writing $u = y^x$, $v = x^y$, $w = x^x$: $$\frac{du}{dx} = y^x\left(\log y + \frac{x}{y}\frac{dy}{dx}\right), \quad \frac{dv}{dx} = x^y\left(\frac{y}{x} + \log x \frac{dy}{dx}\right), \quad \frac{dw}{dx} = x^x(1 + \log x)$$

Step 2 — the right side is constant: the three derivatives sum to $0$.

Step 3 — collect the $\frac{dy}{dx}$ terms: $$\frac{dy}{dx} = -\frac{y^x\log y + y\,x^{y-1} + x^x(1 + \log x)}{x\,y^{x-1} + x^y\log x}$$

Answer: as displayed — each exotic power gets logged separately, then ordinary implicit collection finishes the job.