AP Calculus AB and BC

Derivative of x^(sin x) by Log Differentiation

Q: How do you differentiate x^sin x?

Take logs first. From y = x^{\sin x} you get \ln y = \sin x\ln x, then \frac{y'}{y} = \cos x\ln x+\frac{\sin x}{x}, and multiplying by y gives x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right).

Q: What is the domain of x^sin x?

Take x>0. The log step needs \ln x to exist, and for negative bases a varying real exponent does not give real outputs in general. Nothing at or below zero qualifies, which is why 0.5, 1.2 and 2 are safe values to test.

Logarithmic differentiation is the route that works, and it gives x^(sin x) times (cos x ln x + sin x / x), valid for x > 0. Base and exponent both contain x, so the power rule and the exponential rule are both off the table. Rewriting as e^(sin x ln x) reaches the same answer.

\frac{d}{dx}\left[x^{\sin x}\right] = x^{\sin x}\left(\cos x\ln x + \frac{\sin x}{x}\right)

Neither standard rule applies

The power rule $\frac{d}{dx}x^{n} = nx^{n-1}$ requires a constant exponent. The exponential rule $\frac{d}{dx}a^{x} = a^{x}\ln a$ requires a constant base. In $x^{\sin x}$ the base is $x$ and the exponent is $\sin x$ , so both requirements fail at once.

Applying either rule anyway gives a definite wrong answer, not an approximation. The power rule would give $\sin x\cdot x^{\sin x-1}$ and the exponential rule would give $x^{\sin x}\ln x\cdot\cos x$ . Neither is the derivative on its own, but together they are: $x^{\sin x}\ln x\cos x$ is the first term of the correct answer, and $\sin x\cdot x^{\sin x-1}$ rewrites as $x^{\sin x}\cdot\frac{\sin x}{x}$ , which is the second.

So the correct derivative is exactly the sum of the two wrong answers, and that is no accident: for any function of the form $x^{g(x)}$ the true derivative is $g(x)\,x^{g(x)-1} + x^{g(x)}\ln x\cdot g'(x)$ , one term from each rule. Knowing this is useful as a check. It is not a licence to use either rule alone, since either one on its own is missing half the answer.

Logarithmic differentiation step by step

Name the function $y = x^{\sin x}$ and take the natural log of both sides. The log rule $\ln\left(a^{b}\right) = b\ln a$ pulls the exponent down into a product, which the product rule can handle.

\ln y = \sin x\cdot\ln x

Differentiate both sides with respect to $x$ . The left side needs implicit differentiation, giving $\frac{y'}{y}$ , and the right side is a product of $\sin x$ and $\ln x$ .

\frac{y'}{y} = \cos x\ln x + \sin x\cdot\frac{1}{x}

y' = x^{\sin x}\left(\cos x\ln x + \frac{\sin x}{x}\right)

The mistakes students make

Each of these produces a specific wrong expression that an exam reader will spot immediately.

Answering $\sin x\cdot x^{\sin x-1}$ by treating $\sin x$ as a constant exponent. That expression is one of the two terms of the correct derivative, the $x^{\sin x}\cdot\frac{\sin x}{x}$ piece in disguise, but half an answer is still wrong: at $x=2$ it is about $0.85$ , while the true derivative is about $0.31$ .
Answering $x^{\sin x}\ln x\cos x$ by treating $x$ as a constant base. That is the first term of the correct answer with the entire $\frac{\sin x}{x}$ term missing.
Stopping at $\cos x\ln x + \frac{\sin x}{x}$ . That expression is $\frac{y'}{y}$ , so it still has to be multiplied by $y = x^{\sin x}$ .

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

How do you differentiate x^sin x?

Take logs first. From $y = x^{\sin x}$ you get $\ln y = \sin x\ln x$ , then $\frac{y'}{y} = \cos x\ln x+\frac{\sin x}{x}$ , and multiplying by $y$ gives $x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right)$ .

Can you use the power rule on x^sin x?

No. The power rule only applies when the exponent is a constant. Here the exponent $\sin x$ varies with $x$ , so $\sin x\cdot x^{\sin x-1}$ is not the derivative, although it does turn out to be one of its two terms.

What is the domain of x^sin x?

Take $x>0$ . The log step needs $\ln x$ to exist, and for negative bases a varying real exponent does not give real outputs in general. Nothing at or below zero qualifies, which is why $0.5$ , $1.2$ and $2$ are safe values to test.