How Logarithms Tame a Function Like x to the Power x

The power rule handles $x^5$ and the exponential rule handles $5^x$, but neither handles $x^x$, where both the base and the power vary. Taking logarithms first turns such powers into products. Differentiating twice then shows how a rate itself is changing.

This lesson covers logarithmic differentiation, second-order derivatives, and choosing the right technique.

**For $y=[f(x){]}^{g(x)}$, take natural logarithms to get $\log y=g(x)\log f(x)$, differentiate both sides with respect to x, and multiply by y, giving $\frac{dy}{dx}=y\left[g^{\prime }(x)\log f(x)+\frac{g(x)f^{\prime }(x)}{f(x)}\right]$.

Worked example.** Differentiate $y=x^x$ for $x>0$.

$$\log y=x\log x\Rightarrow \frac{1}{y}\frac{dy}{dx}=\log x+1\Rightarrow \frac{dy}{dx}=x^x(1+\log x)$$

At $x=2$, $\frac{dy}{dx}=4(1+0.693)\approx 6.77$.

Worked example 2. For $y=(\sin x{)}^{\cos x}$:

$$\frac{dy}{dx}=(\sin x{)}^{\cos x}\left[-\sin x\log \sin x+\frac{{\cos }^{2}x}{\sin x}\right]$$

Long products and quotients. For $y=\frac{(x+1{)}^2\sqrt{x-1}}{(x+4{)}^3}$, $\log y=2\log (x+1)+\frac{1}{2}\log (x-1)-3\log (x+4)$, so $\frac{dy}{dx}=y\left[\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}\right]$.

An everyday example. Estimating how fast the number of possible codes grows when a lock uses n positions, each with n symbols, means differentiating $n^n$ — exactly this method.

The substance. **For $x^x$, neither the power rule nor the exponential rule alone is right** — they give $x^x$ and $x^x\log x$, and the true derivative is their sum.

**The second-order derivative $\frac{d^{2}y}{d{x}^2}$ is the derivative of $\frac{dy}{dx}$, found by differentiating the first derivative again, and it measures how fast the slope itself is changing.

Worked example (explicit).** For $y=x^3-6x^2+4$:

$$\frac{dy}{dx}=3x^2-12x,\frac{d^{2}y}{d{x}^2}=6x-12$$

At $x=3$, $\frac{d^{2}y}{d{x}^2}=6$.

Worked example (proving a relation). If $y=e^{2x}+e^{-3x}$, then $y^{\prime }=2e^{2x}-3e^{-3x}$ and $y^{\prime \prime }=4e^{2x}+9e^{-3x}$. Hence $y^{\prime \prime }+y^{\prime }-6y=(4+2-6)e^{2x}+(9-3-6)e^{-3x}=0$.

Worked example (parametric). For $x=a{t}^2$, $y=2at$, $\frac{dy}{dx}=\frac{1}{t}$, so

$$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{1}{t}\right)÷\frac{dx}{dt}=-\frac{1}{t^2}\times \frac{1}{2at}=-\frac{1}{2a{t}^3}$$

An everyday example. A car's position, velocity and acceleration are a function, its first derivative and its second derivative.

The substance. **$\frac{d^{2}y}{d{x}^2}$ is not ${\left(\frac{dy}{dx}\right)}^2$** — and for parametric curves it is not $\frac{d^{2}y/d{t}^2}{d^{2}x/d{t}^2}$ either.

Choose by the structure of the function: the chain rule for one function inside another, logarithmic differentiation for variable powers or long products, implicit differentiation when y is tangled with x, and parametric differentiation when x and y are both given in terms of a third variable.

A quick guide:

- Composite, such as $\sqrt{1+\sin x}$ — chain rule
- Variable power or long product, such as $(1+x{)}^{\tan x}$ — logarithmic differentiation
- Implicit, such as $x^{2}y+y^3=10$ — implicit differentiation
- Parametric, such as $x=t^2$, $y=t^3$ — parametric differentiation
- Sums of variable powers, such as $x^x+(\sin x{)}^x$ — differentiate each term separately

Worked example. Find $\frac{dy}{dx}$ for $y=x^{\sin x}+{\sin }^{-1}\sqrt{x}$.

- $u=x^{\sin x}$ gives $\frac{du}{dx}=x^{\sin x}\left(\cos x\log x+\frac{\sin x}{x}\right)$
- $v={\sin }^{-1}\sqrt{x}$ gives $\frac{dv}{dx}=\frac{1}{\sqrt{1-x}}\times \frac{1}{2\sqrt{x}}$
- $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

Worked example 2. For $x=t^2$ and $y=t^3$, $\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}$, which is 3 at $t=2$.

An everyday example. A mechanic choosing a spanner, a screwdriver or pliers by looking at the fitting makes the same kind of decision as choosing a method from the form of a function.

The substance. Never take the logarithm of a sum in one step — $\log (u+v)$ does not split, so $x^x+(\sin x{)}^x$ must be handled term by term.