How to Differentiate x to the Power x Without Getting Stuck

The power rule handles $x^5$ and the exponential rule handles $5^x$ — but what about $x^x$, where the base and the exponent both change? Curves given through a parameter, and questions about how a rate itself changes, need further tools.

This part covers exponential and logarithmic functions, logarithmic differentiation, parametric derivatives, and second order derivatives.

**The exponential function $e^x$ and the natural logarithm $\log x$ are inverses of each other, with derivatives $\frac{d}{dx}{e}^x=e^x$ and $\frac{d}{dx}\log x=\frac{1}{x}$ for $x>0$.

Exponential function** $y=b^x$ with $b>1$:

- Domain $\mathbb{R}$; range the positive real numbers
- Always increasing, passing through $(0,1)$
- The base $e\approx 2.718$ gives the natural exponential $e^x$

Logarithmic function $y={\log }_{b}x$:

- Domain the positive real numbers; range $\mathbb{R}$
- ${\log }_{b}x=y$ exactly when $b^y=x$
- Laws: $\log (pq)=\log p+\log q$, $\log \frac{p}{q}=\log p-\log q$, $\log p^n=n\log p$

Worked example 1. $\frac{d}{dx}{e}^{3x}=3e^{3x}$, by the chain rule.

Worked example 2. For $y=\log (x^2+1)$:

$$\frac{dy}{dx}=\frac{1}{x^2+1}\cdot 2x=\frac{2x}{x^2+1},{\frac{dy}{dx}|}_{x=1}=\frac{2}{2}=1$$

An everyday example. Money growing with continuously compounded interest follows $e^{rt}$, and its rate of growth at any moment is proportional to the amount already there.

The substance. **$e^x$ equals its own derivative**, which is why it appears whenever a quantity grows at a rate proportional to its size.

**Take the natural logarithm of both sides to turn powers into products and products into sums, differentiate implicitly, and then multiply back by $y$.

When to use it:**

- Functions of the form $[f(x){]}^{g(x)}$, where base and exponent both vary
- Long products or quotients of several factors

**Worked example 1 — $y=x^x$** for $x>0$.

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

$$\frac{dy}{dx}=x^x(1+\log x),{\frac{dy}{dx}|}_{x=1}=1(1+0)=1$$

Worked example 2 — a quotient of products. For $y=\frac{(x+1)(x+2)}{x+3}$:

$$\log y=\log (x+1)+\log (x+2)-\log (x+3)$$

$$\frac{dy}{dx}=y\left[\frac{1}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}\right]$$

At $x=0$, $y=\frac{2}{3}$, so $\frac{dy}{dx}=\frac{2}{3}\left(1+\frac{1}{2}-\frac{1}{3}\right)=\frac{2}{3}\cdot \frac{7}{6}=\frac{7}{9}$.

An everyday example. Tracking how a box's volume changes when its length, width and height all change together is easier after turning the product into a sum of logarithms.

The substance. **Neither the power rule nor the exponential rule works on $x^x$** — one assumes a fixed exponent, the other a fixed base.

**When $x=f(t)$ and $y=g(t)$, differentiate each with respect to $t$ and divide: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$, provided $\frac{dx}{dt}\ne 0$.

Worked example 1 — a circle.** $x=a\cos t$, $y=a\sin t$:

$$\frac{dx}{dt}=-a\sin t,\frac{dy}{dt}=a\cos t,\frac{dy}{dx}=-\frac{\cos t}{\sin t}=-\cot t$$

At $t=\frac{\pi }{4}$, the slope is $-1$.

Worked example 2 — a parabola. $x=a{t}^2$, $y=2at$:

$$\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}$$

With $a=1$ and $t=2$, the point is $(4,4)$ and the slope there is $\frac{1}{2}$.

An everyday example. A cricket ball's horizontal and vertical positions both depend on time, so the direction of its path at any instant is $\frac{dy/dt}{dx/dt}$.

The substance. **There is no need to eliminate $t$** — the parametric formula gives the slope directly, even when eliminating $t$ would be messy.

**The second order derivative $\frac{d^{2}y}{d{x}^2}$ is the derivative of $\frac{dy}{dx}$, and to verify a relation you find $y^{\prime }$ and $y^{\prime \prime }$ and substitute them into the given equation.

Notation.** $\frac{d^{2}y}{d{x}^2}$, $y^{\prime \prime }$, $y_2$ and $f^{\prime \prime }(x)$ all mean the same thing.

Worked example 1. For $y=x^3+2x^2$:

$$y^{\prime }=3x^2+4x,y^{\prime \prime }=6x+4,y^{\prime \prime }(1)=10$$

Worked example 2 — verify a relation. Show that $y=3\cos x+4\sin x$ satisfies $y^{\prime \prime }+y=0$.

$$y^{\prime }=-3\sin x+4\cos x,y^{\prime \prime }=-3\cos x-4\sin x=-y$$

Hence $y^{\prime \prime }+y=0$.

An everyday example. A speedometer shows the first derivative of distance; how quickly that reading climbs when you accelerate is the second derivative.

The substance. **$\frac{d^{2}y}{d{x}^2}$ is not ${\left(\frac{dy}{dx}\right)}^2$** — differentiating twice is different from squaring the first derivative.