Derivative of ln(ax), sin(ax), cos(ax): Chain Rule Explained

Quick answers (for any non-zero constant $a$, $x$ in radians):

$$\frac{d}{dx}\ln(ax) = \frac{1}{x} \qquad \frac{d}{dx}\sin(ax) = a\cos(ax) \qquad \frac{d}{dx}\cos(ax) = -a\sin(ax)$$

The surprising one is the first: the $a$ disappears. This guide shows why, using one chain-rule pattern that covers all three — plus $e^{ax}$, $\tan(ax)$ and more.

The one pattern: $\frac{d}{dx}f(ax) = a\,f'(ax)$

When a function has another function inside it, the chain rule says: differentiate the outside, keep the inside, then multiply by the derivative of the inside.

$$\frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$$

With inside $g(x) = ax$, the inside's derivative is just $a$. So every "$f(ax)$" derivative is the normal derivative of $f$, evaluated at $ax$, multiplied by $a$.

Derivative of ln(ax) = 1/x

Proof 1 — log laws

For $a > 0$ and $x > 0$, $\ln(ax) = \ln a + \ln x$. Since $\ln a$ is a constant, its derivative is 0:

$$\frac{d}{dx}\ln(ax) = 0 + \frac{1}{x} = \frac{1}{x}$$

Proof 2 — chain rule

Outside $\ln(u)$ has derivative $\frac{1}{u}$; inside $u = ax$ has derivative $a$:

$$\frac{d}{dx}\ln(ax) = \frac{1}{ax}\cdot a = \frac{1}{x}$$

The $a$ in the denominator cancels the $a$ from the inside. This works whenever $ax > 0$ — including negative $a$ with negative $x$. For example, for $x < 0$, $\frac{d}{dx}\ln(-2x) = \frac{1}{-2x}\cdot(-2) = \frac{1}{x}$.

The graph that makes it obvious

Because $\ln(ax) = \ln x + \ln a$, the graph of $\ln(5x)$ is just $\ln(x)$ shifted up by $\ln 5 \approx 1.61$. Shifting a curve vertically doesn't change its steepness, so at every $x$ the three curves have exactly the same slope. At $x = 1.5$ each tangent has slope $\frac{1}{1.5} \approx 0.667$.

Example: $\frac{d}{dx}\ln(5x) = \frac{1}{x}$. At $x = 1.3$ the slope is $\frac{1}{1.3} \approx 0.769$ — the same as for $\ln x$.

Derivative of sin(ax) = a cos(ax)

Outside $\sin(u)$ → $\cos(u)$; inside $ax$ → $a$:

$$\frac{d}{dx}\sin(ax) = a\cos(ax)$$

Example: $\frac{d}{dx}\sin(4x) = 4\cos(4x)$. Geometrically, $\sin(4x)$ oscillates 4 times faster than $\sin x$, so its slopes are 4 times steeper.

Derivative of cos(ax) = −a sin(ax)

Outside $\cos(u)$ → $-\sin(u)$; inside → $a$:

$$\frac{d}{dx}\cos(ax) = -a\sin(ax)$$

Example: $\frac{d}{dx}\cos(3x) = -3\sin(3x)$. Don't drop the minus sign — it comes from the derivative of cosine itself.

Cheat sheet

Note the last row: with an added constant $b$, the $a$ no longer cancels completely.

Beyond ax: when the inside is more complicated

The same chain rule handles any inside function — you just multiply by its derivative.

• $\frac{d}{dx}\ln(x^2+1) = \frac{1}{x^2+1}\cdot 2x = \frac{2x}{x^2+1}$
• $\frac{d}{dx}\sin(x^2) = \cos(x^2)\cdot 2x = 2x\cos(x^2)$

Common mistakes

• Answering $\frac{a}{x}$ or $\frac{1}{ax}$ for $\ln(ax)$ — the correct answer is $\frac{1}{x}$.
• Forgetting the factor $a$ for sin, cos and exp.
• Losing the minus sign on the derivative of cosine.
• Working in degrees. These formulas assume radians; in degrees an extra factor of $\frac{\pi}{180}$ appears.

Check with a derivative calculator

Use the derivative of ln calculator, the derivative of cos(ax) calculator, the sin derivative calculator or the exponential derivative calculator to confirm each result step by step. For any other inner function, the chain rule calculator shows the outside/inside split. New to derivatives? Start with the calculus step-by-step guide.

FAQ

What is the derivative of ln(ax)? $\frac{1}{x}$, for any non-zero constant $a$ where $ax > 0$.

Why doesn't the a appear in the derivative of ln(ax)? Because $\ln(ax) = \ln a + \ln x$, and the constant $\ln a$ differentiates to zero. With the chain rule, the $a$ cancels.

What is the derivative of cos(ax)? $-a\sin(ax)$.

What is the derivative of sin(ax)? $a\cos(ax)$.

What is the derivative of e^(ax)? $a\,e^{ax}$.

Do these formulas work in degrees? No — they assume radians. In degrees, multiply by an extra $\frac{\pi}{180}$.

Further reading: Wolfram MathWorld — Chain Rule · OpenStax Calculus Volume 1, chapter 3.