Deriving the Composite Cosine Derivative from First Principles

Why a Changing Input Matters

A cosine function can vary because its input varies. When the input is itself a function of \(x\), the change in that input need not equal the change in \(x\). To find the derivative from first principles, we must keep track of both changes rather than simply replace the input by a more complicated expression.

This tutorial derives the derivative of \(\cos(f(x))\) directly from its difference quotient. The argument will use the cosine addition identity and the standard trigonometric limit \(\lim_{t\to 0}\frac{\sin t}{t}=1\), with angles measured in radians. The result will show exactly where the negative sine factor and the derivative of the inner function come from.

The Difference Quotient for a Composite Cosine

Let \(f\) be differentiable at \(x\), and set \(u=f(x)\). For a nonzero increment \(h\), define the change in the inner input by

Then \(f(x+h)=u+\Delta u\), and the difference quotient for the composite function is

The cosine addition identity states that \(\cos(a+b)=\cos a\cos b-\sin a\sin b\). Applying it with \(a=u\) and \(b=\Delta u\) gives

Therefore the difference quotient separates into two terms:

The first term will approach zero. In the second term, the sine change can be separated into a ratio involving \(\Delta u\) and the rate at which the inner input changes with \(x\). These two facts give a direct derivation.

Derivation from First Principles

Because \(f\) is differentiable at \(x\), the quotient \(\frac{\Delta u}{h}\) approaches the finite value \(f'(x)\) as \(h\to 0\). In particular, this quotient remains bounded for sufficiently small nonzero \(h\). Also, differentiability implies continuity at \(x\), so \(\Delta u\to 0\).

To handle the first term, use the identity \(\cos t-1=-2\sin^2(t/2)\). The standard inequality \(|\sin z|\leq |z|\) then gives

As \(h\to 0\), the factor \(|\Delta u|\) approaches zero and the factor \(\left|\frac{\Delta u}{h}\right|\) remains bounded. Thus the product approaches zero. Multiplication by the fixed value \(\cos u\) does not change that limit.

For the second term, when \(\Delta u\neq 0\), factor its quotient as

As \(h\to 0\), we have \(\Delta u\to 0\). The standard limit \(\lim_{t\to 0}\frac{\sin t}{t}=1\) implies that the first factor approaches \(1\); the definition of \(f'(x)\) implies that the second approaches \(f'(x)\). If \(\Delta u=0\) for some increments, the original sine quotient is zero for those increments, and the same limiting conclusion follows by continuity of \(\frac{\sin t}{t}\) when its value at \(t=0\) is defined to be \(1\). Hence \(\frac{\sin(\Delta u)}{h}\to f'(x)\).

Taking limits in the separated difference quotient now yields the result.

Worked Applications

Why the Two Factors Appear

The derivation has a useful interpretation, even though it did not assume the chain rule. A small change \(h\) in \(x\) produces an inner change \(\Delta u\). The ratio \(\frac{\Delta u}{h}\) describes the inner function's change per unit change in \(x\), and its limit is \(f'(x)\). Meanwhile, the cosine's change per unit change in its input is captured by \(\frac{\sin(\Delta u)}{\Delta u}\), together with the fixed factor \(-\sin u\) from the addition identity.

The negative sign is not optional: it comes from the minus sign in \(\cos(u+\Delta u)=\cos u\cos(\Delta u)-\sin u\sin(\Delta u)\). The inner derivative is also essential: the cosine's input changes at the rate \(f'(x)\), not necessarily at rate \(1\). These observations explain the structure of the formula without treating it as a rule to memorize without justification.

Common Pitfall: Losing the Sign or the Inner Rate

A common incorrect calculation is

This has the correct inner derivative \(2x\), but the sign is wrong. The error often comes from remembering the sine derivative's positive sign rather than using the cosine addition identity. The corrected calculation is

A different mistake is to omit the inner rate and write \(-\sin(x^2)\). That expression differentiates cosine with respect to its input \(u\), but the requested derivative is with respect to \(x\). Since the input \(x^2\) changes at rate \(2x\), the factor \(2x\) must be included. For instance, at \(x=1\), the incorrect omitted-factor answer is \(-\sin(1)\), whereas the difference quotient calculation gives \(-2\sin(1)\).

When This Derivation Applies

The argument requires \(f\) to be differentiable at the point in question. This guarantees both that \(\frac{\Delta u}{h}\) has a finite limit and that \(\Delta u\to0\). It does not require \(f\) to be a polynomial or to have a derivative everywhere: the conclusion is pointwise wherever \(f\) is differentiable. If the inner function is not differentiable at a point, this derivation does not establish differentiability of the composite there.

The radian convention also matters. The standard limit used above has value \(1\) for radian measure. If angles were measured in another unit, the corresponding trigonometric limit would have a different constant, so the derivative formula would need an additional scale factor.