Trigonometric Derivatives: Applying Product, Quotient & Chain Rules

Why does the derivative of $\sin x$ turn out to be $\cos x$, and not some messier expression? And once you know that, how do you handle something like $x^2 \sin x$, or $\sin(x^2)$, where trig functions get tangled up with everything else?

The short answer: sin and cos come from a point moving around a circle, and every "harder" derivative rule (product, quotient, chain) is really just asking how the pieces of a moving system respond to a tiny nudge. Once you see that, you're not memorizing four separate rules. You're applying the same idea four times.

The circle that already knows the answer

Picture a point walking around a circle of radius 1, starting at angle $\theta = 0$ and moving counterclockwise. Its coordinates are $(\cos\theta, \sin\theta)$. That's the definition of sine and cosine, not something we prove, just what those words mean.

Now ask: as $\theta$ increases a tiny bit, which direction does the point move? Not around the circle in general, but specifically: how fast is its height changing, and how fast is its horizontal position changing?

Here's the geometric fact that makes everything else work: the point's velocity is always perpendicular to the line from the center to the point. If the point is at $(\cos\theta, \sin\theta)$, its velocity vector points in the direction $(-\sin\theta, \cos\theta)$. That's not a coincidence you need to memorize, it's just what "moving in a circle" means.

Read off the vertical component of that velocity and you get the rate of change of the height, which is $\sin\theta$. So:

And the horizontal component gives you:

Drag the point below and watch its velocity arrow. Notice it always matches the slope of the sine curve at that exact same instant.

That's the whole foundation. Every trig derivative in this post traces back to those two facts. The hard part isn't sin and cos anymore, it's what happens when they get multiplied, divided, or nested inside other functions.

Two functions glued together: the product rule

Say you need the derivative of $x^2 \sin x$. You can't just take the derivative of $x^2$ and the derivative of $\sin x$ and multiply them, that gives the wrong answer. So what actually happens when two changing quantities get multiplied?

Think of it as a rectangle. One side has length $u(x)$, the other has length $v(x)$, and the area is $u(x) \cdot v(x)$. Now nudge $x$ by a small amount $dx$. Both sides grow a little: $u$ grows by $du$, and $v$ grows by $dv$. The new area is bigger by three pieces:

• A thin strip on the right, width $du$, height $v$: area $v \, du$
• A thin strip on top, width $u$, height $dv$: area $u \, dv$
• A tiny corner square where both strips overlap: area $du \, dv$

The first two strips are the real signal. The corner square is the product of two things that are already small, which makes it smaller still, and it shrinks to nothing as $dx \to 0$. Drag the slider below down toward zero and watch it happen.

Drop that corner term and you're left with the product rule:

For $x^2 \sin x$: let $u = x^2$, so $u' = 2x$, and $v = \sin x$, so $v' = \cos x$. Plug in:

That's it. No new trig facts needed, just the sin and cos derivatives from the circle, combined with the rectangle logic.

Tangent is just a fraction. Fractions need care.

$\tan x$ is defined as $\sin x / \cos x$. So finding its derivative means figuring out how a ratio changes, and that turns out to behave differently from a product.

Start from the product rule in reverse. If $h = f/g$, then $f = h \cdot g$. Differentiate both sides with the product rule:

Solve for $h'$:

That's the quotient rule. Notice the denominator gets squared. That's not decoration, it's what you get from clearing the $g$ you divided by twice: once in the original ratio, once in solving for $h'$.

Apply it to $\tan x = \sin x / \cos x$: here $f = \sin x$ (so $f' = \cos x$) and $g = \cos x$ (so $g' = -\sin x$):

The $g^2$ in the denominator isn't just algebra bookkeeping. Watch what happens as $x$ approaches the point where $\cos x$ shrinks toward zero: the slope of $\tan x$ shoots up, because you're dividing by a number that's getting squared and getting tiny at the same time.

Functions inside functions: the chain rule

What about $\sin(x^2)$? This isn't a product and it isn't a ratio, it's one function stuffed inside another. The outer function is "take the sine," the inner function is "square the input."

Think of it as a linkage: turning an input dial by a small amount turns a second dial connected to it, which in turn turns a third dial connected to that one. If the second dial turns 3 times as fast as the first, and the third turns twice as fast as the second, then the third dial turns $3 \times 2 = 6$ times as fast as the first. Rates through a chain multiply.

That's exactly what happens with $h(x) = f(g(x))$. A small nudge $dx$ in $x$ causes $u = g(x)$ to change by roughly $g'(x)\,dx$. That change in $u$ then causes $y = f(u)$ to change by roughly $f'(u)$ times as much. Multiply the two rates together:

That's the chain rule. Below, $x$ drives $u = x^2$, which drives $y = \sin(u)$. Move $x$ and watch the rate arrow on the middle track get scaled by $g'(x) = 2x$, then the rate arrow on the bottom track get scaled again by $f'(u) = \cos u$.

For $\sin(x^2)$: the outer function is $f(u) = \sin u$, so $f'(u) = \cos u$. The inner function is $g(x) = x^2$, so $g'(x) = 2x$. Multiply:

Same idea for $\cos(3x)$: outer function $\cos u$ gives $-\sin u$, inner function $3x$ gives rate $3$:

When all three show up at once

Real problems don't politely wait for one rule at a time. Take $x \sin(2x)$. This is a product, $u = x$ and $v = \sin(2x)$, so start with the product rule:

But $v = \sin(2x)$ is itself a chain: outer function sine, inner function $2x$. So $v' = \cos(2x) \cdot 2 = 2\cos(2x)$. Substitute that back in:

Notice nothing new happened here. You just used the product rule and the chain rule as tools, one after another, on the pieces that needed them.

The takeaway

Sine and cosine come from a point moving in a circle, and their derivatives are just that point's velocity, read off in two directions. Once you have those two facts, every other trig derivative rule is really a rule about combining changing quantities: multiply two things and their rates add up in a product, divide two things and the rates fight it out with a squared denominator, nest one function inside another and their rates multiply down the chain. You're not learning four unrelated tricks. You're watching the same small idea, a tiny nudge and how it propagates, play out in four different shapes.