Derivative of x^x: Logarithmic Differentiation Explained

The derivative of x^x for x > 0 is x^x(ln x + 1). Both the base and the exponent contain the variable x. So, you cannot directly apply standard power and exponential derivative rules. Instead, take the natural logarithm of both sides. This allows implicit differentiation. Then, you use the product rule to get the final result.

Why neither standard derivative rule works alone

When differentiating basic algebraic expressions, calculus relies on two familiar rules:

• The power rule handles variable bases. It uses constant exponents. For example, d/dx [x^n] = n·x^(n-1).
• The exponential rule handles constant bases. It uses variable exponents. For instance, d/dx [a^x] = a^x·ln a.

The function y = x^x presents a unique challenge. In this function, x appears in both the base and the exponent. Neither value stays constant as x changes. For example, if you use only the power rule, you treat the exponent as constant. This gives you x·x^(x-1) = x^x. Also, if you use only the exponential rule, you treat the base as constant. This gives you x^x·ln x. However, both results are incomplete. Each one ignores half of the rate of change.

You must use logarithmic differentiation to differentiate functions. This applies when both the base and exponent vary. You can also convert the expression into an exponential with base e.

Method 1: logarithmic differentiation step-by-step

Logarithmic differentiation simplifies variable exponents by converting powers into products using logarithm properties. Here is the full derivation for y = x^x when x > 0:

1. Set up the equation: Let y = x^x.
2. Take the natural logarithm of both sides: So, ln y = ln(x^x).
3. Apply the power property of logarithms: Move the exponent to the front to get ln y = x·ln x.
4. Differentiate both sides with respect to x: On the left side, apply the chain rule. The derivative of ln y with respect to x is (1/y)·(dy/dx). On the right side, apply the product rule to x·ln x:

   $$\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}[x]\cdot \ln x + x\cdot \frac{d}{dx}[\ln x]$$

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

5. Isolate dy/dx: Multiply both sides by y to solve for the derivative:

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

6. Substitute back the original function for y: Replace y with x^x to express the final answer solely in terms of x:

   $$\frac{dy}{dx} = x^x(\ln x + 1)$$

Because y represents a function of x, you must differentiate ln y. This requires implicit differentiation. It also requires the chain rule. As a result, you get (1/y)(dy/dx).

Method 2: rewrite with base e

An alternative to logarithmic differentiation is rewriting x^x in terms of the natural exponential function e. Since e^(ln x) = x for all x > 0, express x^x as:

$$x^x = (e^{\ln x})^x = e^{x\ln x}$$

You can differentiate e^(x·ln x). Use the standard exponential chain rule. This rule is d/dx [e^u] = e^u · (du/dx). Here, u = x·ln x.

$$\frac{d}{dx}[x^x] = \frac{d}{dx}[e^{x\ln x}] = e^{x\ln x}\cdot \frac{d}{dx}[x\ln x]$$

You can use the product rule on x·ln x. This gives d/dx [x·ln x] = ln x + 1. Next, substitute this back into the equation.

$$\frac{d}{dx}[x^x] = e^{x\ln x}(\ln x + 1) = x^x(\ln x + 1)$$

Both methods yield identical results. Logarithmic differentiation separates the logarithmic transformation into explicit steps. In contrast, rewriting with base e keeps the expression on a single line.

Why the answer is the sum of two false rules

Distributing x^x into the final derivative reveals an interesting structural pattern:

$$x^x(\ln x + 1) = x^x\cdot 1 + x^x\ln x = x\cdot x^{x-1} + x^x\ln x$$

The term x·x^(x-1) is what you get if you incorrectly treat the exponent as a constant. You would apply the power rule. Also, the term x^x·ln x is what you get if you incorrectly treat the base as a constant. You would apply the exponential rule. However, the true derivative is the sum of these two individual partial calculations.

In multivariable calculus, you find the total derivative of a function. For example, consider z = f(u, v). Both u and v depend on x. You use partial derivatives to find this.

$$\frac{dz}{dx} = \frac{\partial z}{\partial u}\frac{du}{dx} + \frac{\partial z}{\partial v}\frac{dv}{dx}$$

First, set z = u^v. Here, u = x and v = x. This gives partial derivatives. They match the power rule and the exponential rule. Next, add them together. This accounts for both simultaneous changes.

Where is x^x smallest? Critical points and concavity

Finding the local extrema of y = x^x requires setting the first derivative equal to zero to locate critical points:

$$x^x(\ln x + 1) = 0$$

Since x^x > 0 for all real numbers x > 0, the equation can only equal zero when:

$$\ln x + 1 = 0 \implies \ln x = -1 \implies x = e^{-1} = \frac{1}{e} \approx 0.3679$$

Evaluating f(1/e) gives the minimum height:

$$f\left(\frac{1}{e}\right) = \left(\frac{1}{e}\right)^{1/e} = e^{-1/e} \approx 0.6922$$

To verify this critical point as a local minimum, compute the second derivative. Use the product rule on x^x(ln x + 1).

$$\frac{d^2}{dx^2}[x^x] = \frac{d}{dx}[x^x](\ln x + 1) + x^x\cdot \frac{d}{dx}[\ln x + 1]$$

$$\frac{d^2}{dx^2}[x^x] = x^x(\ln x + 1)^2 + x^x\cdot \frac{1}{x} = x^x(\ln x + 1)^2 + x^{x-1}$$

For all x > 0, both x^x(ln x + 1)^2 and x^(x-1) are strictly positive. Also, the second derivative is strictly positive for all x > 0. Therefore, the graph of y = x^x is concave upward across its entire domain. This makes x = 1/e a global minimum on (0, ∞). You can check your calculations using the critical points finder.

Behavior near zero and the limit of x^x

The expression 0^0 is an indeterminate form. You cannot evaluate x^x directly at x = 0. However, the right-sided limit as x approaches zero yields a defined value.

$$\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x\ln x}$$

You need to evaluate the limit of the exponent x·ln x as x → 0^+. Rewrite it as a quotient. This allows you to apply L’Hôpital’s rule.

$$\lim_{x \to 0^+} x\ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$$

Therefore, the overall limit is:

$$\lim_{x \to 0^+} x^x = e^0 = 1$$

However, examining the limit of the derivative as x → 0^+ reveals contrasting behavior:

$$\lim_{x \to 0^+} x^x(\ln x + 1) = 1 \cdot (-\infty) = -\infty$$

As x approaches 0 from the right, the function height approaches 1. However, the tangent line becomes infinitely steep in the negative direction. The graph enters from (0, 1) with an almost vertical downward curve. It bottoms out at x = 1/e ≈ 0.3679. Then it grows rapidly.

Worked examples of variable exponents

Logarithmic differentiation applies to a wide range of functions with variable bases and variable exponents:

1. Slope and tangent line of y = x^x at x = 1:
Substitute x = 1 into the derivative: dy/dx = 1^1(ln 1 + 1) = 1(0 + 1) = 1. At point (1, 1), the slope is 1. The equation of the tangent line is y – 1 = 1(x – 1), which simplifies to y = x.

2. Slope and tangent line of y = x^x at x = 2:
Substitute x = 2 into the derivative: dy/dx = 2^2(ln 2 + 1) = 4(ln 2 + 1) ≈ 6.7726. At point (2, 4), the tangent line equation is y – 4 = 4(ln 2 + 1)(x – 2).

3. Find the derivative of y = x^(sin x):
First, take natural logs. So, ln y = sin x · ln x.
Next, differentiate using the product rule. This gives you (1/y)·y’ = cos x · ln x + sin x · (1/x).
Finally, multiply by y = x^(sin x). Therefore, y’ = x^(sin x) [cos x · ln x + (sin x)/x].

Derivative of x^(1/x)

4. Find the derivative of y = x^(1/x) and locate its maximum:
Take natural logs. So, ln y = (ln x)/x.
Differentiate using the quotient rule. Thus, (1/y)·y’ = \[ (1/x)·x - \ln x · 1 \] / x^2. This simplifies to (1 – ln x) / x^2.
Multiply by y = x^(1/x). Therefore, y’ = x^(1/x) · (1 – ln x) / x^2.
Set y’ = 0. This gives 1 – ln x = 0. So, x = e. The function achieves its absolute maximum at x = e. Here, e^(1/e) ≈ 1.4447.

5. Find the derivative of y = x^(x^2):
First, take natural logs. So, ln y = x^2 · ln x.
Next, differentiate using the product rule. This yields (1/y)·y’ = 2x · ln x + x^2 · (1/x). Thus, you get 2x · ln x + x.
Finally, multiply by y = x^(x^2). So, y’ = x^(x^2) (2x · ln x + x).

6. Find the derivative of y = (ln x)^x for x > 1:
First, take natural logs. So, ln y = x · ln(ln x).
Next, differentiate using the product and chain rules. This gives you (1/y)·y’ = 1 · ln(ln x) + x · [1 / ln x] · (1/x). Therefore, you have ln(ln x) + 1 / ln x.
Finally, multiply by y = (ln x)^x. So, y’ = (ln x)^x \[ \ln(\ln x) + 1 / \ln x \].

General formula for derivatives of the form f(x)^g(x)

You can establish a general formula for any function y = [f(x)]^[g(x)]. This applies when f(x) > 0. Also, both f and g must be differentiable.

1. Take the natural log: ln y = g(x) · ln[f(x)]
2. Differentiate implicitly:

   $$\frac{1}{y}\frac{dy}{dx} = g'(x)\ln[f(x)] + g(x)\cdot \frac{f'(x)}{f(x)}$$

3. Multiply by y = [f(x)]^[g(x)]:

   $$\frac{d}{dx}\left[[f(x)]^{g(x)}\right] = [f(x)]^{g(x)}\left( g'(x)\ln[f(x)] + \frac{g(x)f'(x)}{f(x)} \right)$$

The term g'(x)ln[f(x)] handles the rate of change from the exponent. Also, g(x)f'(x)/f(x) handles the rate of change from the base.

Comparing x², 2^x, and x^x

Choosing the appropriate derivative rule depends on identifying where the variable appears in an exponential expression:

For large values of x, x^x grows drastically faster than algebraic polynomials. For example, it grows faster than x^2. It also grows faster than standard exponential functions like 2^x. This is because both the base and the exponent increase simultaneously.

Using logarithmic differentiation for complex products

Logarithmic differentiation is also an effective technique for simplifying rational expressions involving multiple products, quotients, and powers:

$$y = \frac{x^3(x+1)^4}{x^2+1}$$

Taking the natural logarithm breaks multiplication into addition and division into subtraction:

$$\ln y = \ln[x^3(x+1)^4] - \ln(x^2+1) = 3\ln x + 4\ln(x+1) - \ln(x^2+1)$$

Differentiating both sides with respect to x yields:

$$\frac{1}{y}\frac{dy}{dx} = \frac{3}{x} + \frac{4}{x+1} - \frac{2x}{x^2+1}$$

Multiplying both sides by y completes the solution:

$$\frac{dy}{dx} = \left(\frac{x^3(x+1)^4}{x^2+1}\right)\left( \frac{3}{x} + \frac{4}{x+1} - \frac{2x}{x^2+1} \right)$$

Common mistakes to avoid

• Applying only the power rule:
  Writing d/dx[x^x] = x·x^(x-1) = x^x treats the exponent as constant. However, this omits x^x·ln x.
• Applying only the exponential rule: You write d/dx[x^x] = x^x·ln x. However, this treats the base as constant. It also omits the power-rule contribution.
• Forgetting to multiply by y at the end: You differentiate ln y. This gives you (1/y)(dy/dx). Then, you must multiply by y = x^x. This isolates dy/dx.
• Incorrectly applying the product rule: Remember that d/dx[x] = 1 when you differentiate x·ln x. This gives you 1·ln x + x·(1/x). So, the result is ln x + 1.
• Overlooking domain constraints: Logarithmic differentiation needs a positive argument inside the natural logarithm. Therefore, restrict x > 0 when you work with x^x.

Try it yourself

For additional functions with variable exponents, enter them into the derivative calculator to verify your steps.

Practice problems

Test your understanding of logarithmic differentiation on these practice problems:

1. Find the derivative of f(x) = x^(2x).
2. Next, find the derivative of f(x) = x^(ln x).
3. Find the derivative of f(x) = (sin x)^x on the interval (0, π).
4. Also, find the derivative of f(x) = x^(√x).
5. Find the derivative of f(x) = (x^2 + 1)^x.
6. Calculate the exact slope of y = x^x at x = 2.

Solutions:
(1) x^(2x)(2 ln x + 2)
(2) x^(ln x) · (2 ln x / x)
(3) (sin x)^x [ln(sin x) + x cot x]
(4) x^(√x) [(ln x)/(2√x) + 1/√x]
(5) (x^2 + 1)^x [ln(x^2 + 1) + (2x^2)/(x^2 + 1)]
(6) 4 + 4 ln 2

Frequently asked questions

Is the derivative of x^x equal to x·x^(x−1)?

No, you cannot apply x·x^(x-1). This assumes that the exponent is a constant number. It yields only x^x. As a result, it misses the second term, x^x·ln x.

Can you differentiate x^x for negative x values?

For most negative real numbers, x^x produces non-real complex numbers, making the standard real derivative undefined. Calculus restricts the domain of x^x to x > 0.

What is the second derivative of x^x?

The second derivative of x^x is x^x(ln x + 1)^2 + x^(x-1). You derive this by taking the derivative of x^x(ln x + 1). You must use the product rule.

What is the integral of x^x?

There is no closed-form antiderivative for x^x using elementary functions. Therefore, you must compute definite integrals using numerical integration. Alternatively, you can use series expansions. For example, the Sophomore’s dream identity states that the integral from 0 to 1 of x^x dx equals approximately 0.7834.

Why is natural log (ln) used instead of log base 10?

We use natural logarithms. This is because the derivative of ln y is simply (1/y)·y’. It has no extra scaling constants. See derivative of ln x for details.

Further reading

• Logarithmic Differentiation (Paul’s Online Math Notes) — detailed guide and additional worked examples covering variable base and exponent problems.
• Derivatives of Exponential and Logarithmic Functions (OpenStax Calculus Volume 1) — comprehensive textbook exposition on logarithmic differentiation and exponent rules.
• Derivatives of Exponential and Logarithm Functions (Paul’s Online Math Notes) — foundational rules for differentiating logarithmic and exponential functions.