Composition and Trigonometric Derivatives

If a quantity depends on temperature and temperature depends on time, its time rate should reflect both changes. The chain rule makes that statement exact. If $g$ is differentiable at $x$ and $f$ at $g(x)$, then

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

The outer derivative is evaluated at the actual inner value. Writing $f'(x)g'(x)$ usually differentiates a different expression.

A justification that handles zero inner changes

Differentiability gives $f(u+k)-f(u)=f'(u)k+k\eta(k)$, where $\eta(k)\to0$ as $k\to0$ and we may set $\eta(0)=0$. Put $u=g(x)$ and $k=g(x+h)-g(x)$. Divide by $h$, not by $k$:

$$\frac{f(g(x+h))-f(g(x))}h =[f'(g(x))+\eta(k)]\frac{g(x+h)-g(x)}h.$$

Continuity of $g$ gives $k\to0$, and the second quotient tends to $g'(x)$. The product has the claimed limit. This avoids the common gap in a proof that divides by $g(x+h)-g(x)$ even though that difference can be zero for arbitrarily small nonzero $h$.

For $F(x)=(3x^2+1)^4$, the outer function is $u^4$ and the inner is $3x^2+1$. Hence $F'=4(3x^2+1)^3(6x)$. For $\sqrt{1+x^2}$, the result is $x/\sqrt{1+x^2}$. Its denominator is always positive, so the derivative exists for all real $x$.

Sine from first principles

The angle-addition identity gives

$$\frac{\sin(x+h)-\sin x}h =\sin x\frac{\cos h-1}h+\cos x\frac{\sin h}h.$$

The two trigonometric limits previously established are $0$ and $1$. Therefore $(\sin x)'=\cos x$. Applying the cosine angle-addition identity similarly yields $(\cos x)'=-\sin x$. These formulas assume radians. They were not used to prove the limits on which their derivation relies.

The quotient rule now gives

$$(\tan x)'=\frac{\cos^2x+\sin^2x}{\cos^2x}=\sec^2x,$$

where $\cos x\ne0$. Reciprocals and quotients also give $(\cot x)'=-\csc^2x$, $(\sec x)'=\sec x\tan x$, and $(\csc x)'=-\csc x\cot x$ on their domains. Stating a derivative formula does not remove the poles of the original function.

A time-dependent oscillation

For a modeled displacement $s(t)=2\sin(3t+\pi/4)$ metres, where $t$ is in seconds and the phase is in radians, the velocity is $v(t)=6\cos(3t+\pi/4)$ metres per second. Acceleration is $a(t)=-18\sin(3t+\pi/4)$ metres per second squared. The factor $3$ has units radians per second, with radians dimensionless in the derivative convention. Forgetting the inner derivative loses the frequency factor in every rate.

If an input $\theta$ is explicitly measured numerically in degrees, the function is $\sin(\pi\theta/180)$ in radian notation. Its derivative with respect to the degree-valued input is $(\pi/180)\cos(\pi\theta/180)$. The familiar cosine-only rule is not unit-independent.

Nested operations

For $H(x)=\sin((x^2+1)^3)$, apply the chain rule at each level:

$$H'(x)=\cos((x^2+1)^3)\,3(x^2+1)^2\,2x.$$

For $J(x)=x^2\sin(3x)$, multiplication is outside the composition, so first use the product rule: $J'=2x\sin(3x)+3x^2\cos(3x)$. A derivative can require several rules, each corresponding to a different layer of the expression.

Exercises

1. Differentiate $(1-2x)^7$.
2. Differentiate $\cos(x^2)$ and $\sin^2x$. Explain the different compositions.
3. Find the derivative of $\tan(4x)$ and state where it exists.
4. Differentiate $x\cos x$ twice.
5. A displacement is $5\cos(2t)$ metres. Find velocity and acceleration at $t=\pi/4$ seconds.
6. Explain why a chain-rule proof must not assume every inner increment is nonzero.

Worked solutions

1. $-14(1-2x)^6$.
2. $-2x\sin(x^2)$ and $2\sin x\cos x$. The first applies cosine to a square; the second squares a sine.
3. $4\sec^2(4x)$ wherever $\cos(4x)\ne0$.
4. First derivative $\cos x-x\sin x$; second $-2\sin x-x\cos x$.
5. $v=-10\sin(2t)$ and $a=-20\cos(2t)$, giving $-10$ metres per second and zero acceleration at the stated time.
6. An inner function may be constant locally or return to its base value at many nearby inputs. The remainder formulation does not divide by that increment.