Examples on Chain Rule

This section is pure practice at recognising the outer function / inner function split that the chain rule (Theorem 10.5) needs, and at combining it with the other rules when several layers or several factors are present.

Worked illustration — a square root of a sum. For $F(x)=\sqrt{x^2+1}$, take the inner function $u=g(x)=x^2+1$ and the outer function $f(u)=\sqrt{u}$. Since $f'(u)=\tfrac12 u^{-1/2}=\tfrac{1}{2\sqrt u}$ and $g'(x)=2x$, the chain rule gives $F'(x)=f'(g(x))\,g'(x) = \dfrac{1}{2\sqrt{x^2+1}}\cdot2x = \dfrac{x}{\sqrt{x^2+1}}$.

Worked illustration — telling $\sin(x^2)$ apart from $\sin^2x$. These look similar but have different inner/outer splits. For $y=\sin(x^2)$: inner $u=x^2$, outer $\sin u$, so $\dfrac{dy}{dx}=\cos(u)\cdot2x = 2x\cos(x^2)$. For $y=\sin^2x=(\sin x)^2$: inner $u=\sin x$, outer $u^2$, so $\dfrac{dy}{dx}=2u\cdot\cos x = 2\sin x\cos x = \sin2x$ (using the double-angle identity). The lesson: always ask which operation is applied last — that is the outer function.

Worked illustration — a high power. For $y=(x^3-1)^{100}$: inner $u=x^3-1$ ($u'=3x^2$), outer $u^{100}$ ($100u^{99}$), so $\dfrac{dy}{dx}=100(x^3-1)^{99}\cdot3x^2 = 300x^2(x^3-1)^{99}$ — the chain rule turns what would be an impossible binomial expansion into a one-line answer.

Worked illustration — chain rule nested inside a product. For $y=(2x+1)^5(x^3-x+1)^4$, apply the product rule with $u=(2x+1)^5$ and $v=(x^3-x+1)^4$, computing $u'$ and $v'$ themselves by the chain rule: $u'=5(2x+1)^4\cdot2=10(2x+1)^4$ and $v'=4(x^3-x+1)^3(3x^2-1)$; then $\dfrac{dy}{dx}=uv'+vu' = 4(x^3-x+1)^3(3x^2-1)(2x+1)^5+10(2x+1)^4(x^3-x+1)^4$, which factors to $2(2x+1)^4(x^3-x+1)^3\big[(2x+1)(3x^2-1)\cdot 2+10(x^3-x+1)\big]$ after collecting the common factors $(2x+1)^4(x^3-x+1)^3$ — the chain rule and product rule work together seamlessly.

Worked illustration — exponential composite. For $y=e^{\sin x}$: inner $u=\sin x$ ($u'=\cos x$), outer $e^u$ ($e^u$), so $\dfrac{dy}{dx}=e^u\cos x = \cos x\, e^{\sin x}$. …