Chain Rule Practice

These questions follow The Chain Rule — brackets to a power, then the rational and radical disguises, then chains in the abstract. Before each derivative, say the inside and the outside out loud; the sorting is most of the skill.

The rule itself

1. Differentiate $f(x)=(x^2+1{)}^2$ twice: once with the chain rule, once by expanding first. Confirm the answers agree.
2. Differentiate $f(x)=(2x^3-5{)}^4$.
3. Differentiate $f(x)=(5x^3{)}^{1/3}$ with the chain rule, then simplify the original function first and differentiate again. Confirm the answers agree.

Answer 1

Chain rule: outside is a square, inside is $x^2+1$, so $f^{\prime }(x)=2(x^2+1)\cdot 2x=4x(x^2+1)=4x^3+4x$. Expanding first: $f(x)=x^4+2x^2+1$, so $f^{\prime }(x)=4x^3+4x$. ✓ Same answer, and the expanded road will stop being available the moment the exponent is $\frac{1}{2}$.

Answer 2

Outside: fourth power. Inside: $2x^3-5$, whose derivative is $6x^2$. $f^{\prime }(x)=4(2x^3-5{)}^3\cdot 6x^2=24x^2(2x^3-5{)}^3$ No expansion required — that is the whole point.

Answer 3

Chain rule: $f^{\prime }(x)=\frac{1}{3}(5x^3{)}^{-2/3}\cdot 15x^2=5x^2(5x^3{)}^{-2/3}$, which simplifies to $5^{1/3}$ (the $x^2$ cancels against $x^{-2}$). Simplifying first: $f(x)=5^{1/3}x$, a line, so $f^{\prime }(x)=5^{1/3}$ immediately. Both roads give the constant $5^{1/3}$ — the function was a straight line in disguise, and the chain rule saw through it.

Rational and radical disguises

4. Differentiate $f(x)=\sqrt{x^2+5}$, and evaluate $f^{\prime }(2)$.
5. Express $f(x)=\frac{x^2+1}{x-1}$ as a product, and differentiate it. Write the answer as a single fraction.
6. Differentiate $f(x)=\frac{1}{\sqrt{3x+1}}$, and evaluate $f^{\prime }(1)$.

Answer 4

As a power: $f(x)=(x^2+5{)}^{1/2}$, so $\begin{array}{rl}{f}^{\prime }(x)& =\frac{1}{2}(x^2+5{)}^{-1/2}\cdot 2x\\ & =\frac{x}{\sqrt{x^2+5}}\end{array}$ At $x=2$: $f^{\prime }(2)=\frac{2}{\sqrt{9}}=\frac{2}{3}$.

Answer 5

$f(x)=(x^2+1)(x-1{)}^{-1}$. Product rule, with the chain rule on the second factor: $f^{\prime }(x)=2x(x-1{)}^{-1}+(x^2+1)(-1)(x-1{)}^{-2}$ Common denominator $(x-1{)}^2$: $\begin{array}{rl}{f}^{\prime }(x)& =\frac{2x(x-1)-(x^2+1)}{(x-1{)}^2}\\ & =\frac{x^2-2x-1}{(x-1{)}^2}\end{array}$ Audit at $x=0$: $f^{\prime }(0)=\frac{-1}{1}=-1$, and nearby values of $f$ confirm a slope near $-1$ there.

Answer 6

As a power: $f(x)=(3x+1{)}^{-1/2}$, so $\begin{array}{rl}{f}^{\prime }(x)& =-\frac{1}{2}(3x+1{)}^{-3/2}\cdot 3\\ & =\frac{-3}{2(3x+1{)}^{3/2}}\end{array}$ At $x=1$: $(4{)}^{3/2}=8$, so $f^{\prime }(1)=-\frac{3}{16}$. Negative is right — the function shrinks as its denominator grows.

Chains in the abstract

7. Suppose $h(x)=f(g(x))$, with $g(2)=3$, $g^{\prime }(2)=4$, and $f^{\prime }(3)=5$. Determine $h^{\prime }(2)$, and explain the answer in terms of rates.

Answer 7

$h^{\prime }(2)=f^{\prime }(g(2))\cdot g^{\prime }(2)=f^{\prime }(3)\times 4=5\times 4=20$. In words: near $x=2$ the inner machine runs at 4 output units per input unit, and near the value 3 the outer machine amplifies whatever it receives 5 times over. Rates through a chain multiply: $20$. Note which numbers went unused — $f^{\prime }(2)$ never appeared, because the outer machine is evaluated where the inner one delivers, not where you started.