Corrected exercises: the chain rule (MATH 140)

a) Let $u = 3x^2 - 5$, so $y = u^7$. The outer function is the seventh power, whose derivative is $7u^6$; the inner derivative is $u' = 6x$. By the chain rule, $y' = 7(3x^2 - 5)^6 \cdot 6x = 42x(3x^2 - 5)^6$. At $x = 1$: $u = -2$ and $(-2)^6 = 64$, so $y'(1) = 42 \cdot 64 = 2688$. Expanding $(3x^2 - 5)^7$ first would take a page and invite ten sign errors: the chain rule exists precisely so that the bracket is never opened. The inside stays as it is in the outer derivative, $(3x^2 - 5)^6$, and never becomes $(6x)^6$.

b) Let $u = x^3 + 1$, so $y = \sqrt u = u^{1/2}$ and $\frac{dy}{du} = \frac{1}{2\sqrt u}$. With $u' = 3x^2$: $y' = \frac{1}{2\sqrt{x^3+1}} \cdot 3x^2 = \frac{3x^2}{2\sqrt{x^3 + 1}}$. The rule $(\sqrt u)' = \frac{u'}{2\sqrt u}$ is worth knowing by heart, with its $u'$ on top: the answer $\frac{1}{2\sqrt{x^3+1}}$, without the $3x^2$, is the single most common error of the chapter.

c) Read the order of operations as a calculator would perform it. In $\sin(x^2)$ you square first and take the sine LAST: the outer function is $\sin$, the inner one $u = x^2$, so $y_1' = \cos(x^2) \cdot 2x$. In $\sin^2 x = (\sin x)^2$ you take the sine first and square LAST: the outer function is the square, $u = \sin x$, so $y_2' = 2\sin x \cdot \cos x$. In $x^2\sin x$ the last operation is a PRODUCT, not a composition, so the product rule applies: $y_3' = 2x\sin x + x^2\cos x$. Three expressions built from the same pieces, three different rules: the last operation decides, and a student who reads $\sin^2 x$ as $\sin(x^2)$ loses the whole question.

d) For $y = e^{-x^2/2}$, let $u = -\frac{x^2}{2}$, so $y = e^u$ and $\frac{dy}{du} = e^u$; with $u' = -x$, $y' = -x e^{-x^2/2}$. For $y = e^{1/x}$, let $u = \frac{1}{x} = x^{-1}$, so $u' = -\frac{1}{x^2}$ and $y' = -\frac{e^{1/x}}{x^2}$. The exponential is its own derivative only when its exponent is $x$ itself: with any other exponent, $e^u$ comes back unchanged AND multiplied by $u'$.

e) Let $u = x^2 + 4$, so $y = u^{-3}$ and $y' = -3u^{-4} \cdot u' = -3(x^2 + 4)^{-4} \cdot 2x = -\frac{6x}{(x^2 + 4)^4}$. The student differentiated the outer layer correctly and stopped: the factor $2x$ is missing, and the answer is wrong for every $x \ne \frac{1}{2}$. By the quotient rule, $\left(\frac{1}{v}\right)' = -\frac{v'}{v^2}$ with $v = (x^2+4)^3$, and $v'$ itself needs the chain rule: $v' = 3(x^2+4)^2 \cdot 2x$. So $-\frac{6x(x^2+4)^2}{(x^2+4)^6} = -\frac{6x}{(x^2+4)^4}$, the same answer after a simplification. Rewriting a constant over a power as a negative power is the faster road, and it removes one place to make a mistake.

a) $\sin^3(4x) = (\sin(4x))^3$. Layers, from the outside in: the cube, the sine, the multiplication by $4$. One factor per layer: $3(\sin 4x)^2$ for the cube, $\cos(4x)$ for the sine, $4$ for the inner layer. So $y' = 3\sin^2(4x) \cdot \cos(4x) \cdot 4 = 12\sin^2(4x)\cos(4x)$. Three layers, three factors. The usual loss is the last one: $3\sin^2(4x)\cos(4x)$ looks finished and is off by a factor $4$ everywhere.

b) Layers: the square root, then $1 + w^2$, then $w = \cos x$. Factors: $\frac{1}{2\sqrt{1+\cos^2 x}}$, then $2\cos x$, then $-\sin x$. So $y' = \frac{1}{2\sqrt{1 + \cos^2 x}} \cdot 2\cos x \cdot (-\sin x) = -\frac{\sin x \cos x}{\sqrt{1 + \cos^2 x}}$. The derivative of the constant $1$ is $0$, so the middle layer contributes only $2\cos x$; the minus sign comes from the innermost layer and must survive to the end.

c) Layers: $e^{(\cdot)}$, then $\sin$, then $x^2$. Factors: $e^{\sin(x^2)}$ (the exponential comes back unchanged, with its whole exponent), $\cos(x^2)$, $2x$. So $y' = e^{\sin(x^2)} \cdot \cos(x^2) \cdot 2x$. Check the middle factor: the sine was applied to $x^2$, so its derivative is evaluated at $x^2$ too, $\cos(x^2)$ and not $\cos x$. Each outer derivative is evaluated at exactly what that layer received.

d) Layers: the fifth power, then $1 + \sin w$, then $w = \frac{\pi x}{6}$. So $h'(x) = 5\left(1 + \sin\frac{\pi x}{6}\right)^4 \cdot \cos\frac{\pi x}{6} \cdot \frac{\pi}{6}$. At $x = 1$: $\sin\frac{\pi}{6} = \frac{1}{2}$ and $\cos\frac{\pi}{6} = \frac{\sqrt 3}{2}$, so $h'(1) = 5 \cdot \left(\frac{3}{2}\right)^4 \cdot \frac{\sqrt 3}{2} \cdot \frac{\pi}{6} = 5 \cdot \frac{81}{16} \cdot \frac{\sqrt 3\, \pi}{12} = \frac{405\sqrt 3\,\pi}{192} = \frac{135\sqrt 3\, \pi}{64}$. Differentiate in general FIRST, then substitute $x = 1$: substituting first turns the function into the constant $\left(\frac{3}{2}\right)^5$, whose derivative is $0$.

e) Three errors. First, the derivative of the outer sine is $\cos$ evaluated at what the sine received, $\cos(x^2)$, untouched: the student differentiated INSIDE the cosine instead of multiplying by the inner derivative. Second, the inner layers were differentiated as if $\cos(x^2)$ had derivative $-\sin(2x)$: the cosine layer gives $-\sin(x^2)$, and the square gives a separate factor $2x$, never $\sin(2x)$. Third, a product of factors became a single nested expression. Correctly, with layers $\sin$, $\cos$, $x^2$: $y' = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot 2x = -2x\sin(x^2)\cos(\cos(x^2))$. The chain rule MULTIPLIES factors side by side; it never places one derivative inside another.

a) $h'(1) = f'(g(1)) \cdot g'(1)$. Read $g(1) = 2$ first, then $f'$ at $2$: $f'(2) = 5$. With $g'(1) = 4$, $h'(1) = 5 \cdot 4 = 20$. The student's $8$ is $f'(1) \cdot g'(1) = 2 \cdot 4$: the outer derivative was read at $x = 1$ instead of at $g(1) = 2$. The outer function never sees $x$, only what $g$ hands it, so its slope is read at $g(1)$. That is the whole point of the table question, and a wrong row gives $0$ out of the method marks.

b) $k'(1) = g'(f(1)) \cdot f'(1)$. Read $f(1) = 3$, then $g'(3) = 6$; with $f'(1) = 2$, $k'(1) = 12$. For $G = g^3$, the outer function is the cube and the inner one $g$: $G'(x) = 3[g(x)]^2 g'(x)$, so $G'(2) = 3 \cdot 3^2 \cdot (-1) = -27$. The power of a function is a composition, and $3[g(2)]^2$ alone, without $g'(2)$, is the missing-factor error again.

c) $F'(x) = f'(x^2 - 1) \cdot 2x$. At $x = 2$: $x^2 - 1 = 3$, $f'(3) = -2$, and the inner derivative is $2 \cdot 2 = 4$, so $F'(2) = -8$. The trap is to read $f'(2)$ because the question says $x = 2$: the table is read at the inside, $3$. For $H = f \circ f$: $H'(1) = f'(f(1)) \cdot f'(1) = f'(3) \cdot 2 = -4$. A function composed with itself is still two layers, and its derivative is still two factors read at two different points.

d) On the graph, $f$ has slope $\frac{5 - 1}{2 - 0} = 2$ on $(0, 2)$ and $\frac{3 - 5}{6 - 2} = -\frac{1}{2}$ on $(2, 6)$; $g$ has slope $-1$ on $(0, 3)$ and $1$ on $(3, 6)$. Now $u'(1) = f'(g(1))\, g'(1)$: $g(1) = 3$, which lies in $(2, 6)$ where $f' = -\frac{1}{2}$, and $g'(1) = -1$, so $u'(1) = \frac{1}{2}$. Similarly $g(5) = 3$ and $g'(5) = 1$, so $u'(5) = -\frac{1}{2}$. The same inner VALUE, $3$, gives the same outer slope; the inner SLOPE changes sign, and so does $u'$.

e) $v'(1) = g'(f(1))\, f'(1)$ requires $g'(3)$, and $g$ has a corner at $3$: the chain rule cannot be applied, since its hypothesis, $g$ differentiable at $f(1)$, fails. Look at the one-sided behaviour instead. Just left of $x = 1$, $f(x)$ is slightly below $3$ (because $f$ increases with slope $2$), where $g$ has slope $-1$: $v$ has slope $-1 \cdot 2 = -2$. Just right of $1$, $f(x) > 3$ and $g$ has slope $1$: $v$ has slope $2$. The one-sided slopes differ, so $v'(1)$ does not exist; $v$ has a corner at $1$. For $w$: $g(4) = 2$, $g'(2) = -1$ and $g'(4) = 1$, so $w'(4) = -1$. Checking the hypotheses of the chain rule is part of the answer.

a) Since $b = e^{\ln b}$, $b^x = \left(e^{\ln b}\right)^x = e^{x\ln b}$. Let $u = x\ln b$, so $b^x = e^u$. The outer function is $e^u$, its own derivative; the inner function is a constant times $x$, so $u' = \ln b$. By the chain rule, $\frac{d}{dx}b^x = e^{x \ln b} \cdot \ln b = b^x \ln b$. For $b = e$, $\ln e = 1$ and the familiar $(e^x)' = e^x$ comes back: $e$ is the only base with no extra factor. Writing $\ln b$ as a constant, and saying so, is what the marker checks here.

b) $3^{x^2}$: the outer function is $3^u$, of derivative $3^u \ln 3$, and $u = x^2$: $\frac{d}{dx}3^{x^2} = 3^{x^2} \ln 3 \cdot 2x$. $5^{\sin x}$: $5^{\sin x}\ln 5 \cdot \cos x$. $10^{-x}$: $10^{-x} \ln 10 \cdot (-1) = -\ln 10 \cdot 10^{-x}$. Each answer has the exponential unchanged, the constant $\ln b$, and the derivative of the exponent: three factors, and the most often forgotten is the last.

c) The slope at $0$ of $y = b^x$ is $b^0 \ln b = \ln b$. For $e^x$ it is $\ln e = 1$, exactly the slope of $y = 1 + x$, which also passes through $(0, 1)$: the line is the tangent to $y = e^x$. For $2^x$ the slope is $\ln 2 \approx 0.69 < 1$, and for $3^x$ it is $\ln 3 \approx 1.10 > 1$. On the figure, the dashed line therefore CROSSES $y = 2^x$ and $y = 3^x$ at $(0, 1)$, passing from one side to the other, while it stays below $y = e^x$ on both sides. This is Stewart's definition of $e$ read on a graph: the base whose exponential has slope exactly $1$ at $0$.

d) $\frac{d}{dx}4^x = 4^x \ln 4 = 4^x \cdot 2\ln 2$. Setting $4^x \cdot 2\ln 2 = 8\ln 2$ gives $4^x = 4$, so $x = 1$ and the point is $(1, 4)$. For $8^x$: directly, $(8^x)' = 8^x \ln 8$. As $2^{3x}$, the outer function is $2^u$ with $u = 3x$: $(2^{3x})' = 2^{3x} \ln 2 \cdot 3$. The two agree because $2^{3x} = 8^x$ and $3\ln 2 = \ln 2^3 = \ln 8$, a law of logarithms, not a derivative. Two correct routes must meet, and checking that they do is the cheapest verification there is.

e) $x^\pi$ is a power of $x$ with a constant exponent: power rule, $\pi x^{\pi - 1}$. $\pi^x$ is an exponential with a constant base: $\pi^x \ln \pi$. $\pi^\pi$ is a constant: its derivative is $0$. The student applied the power rule to an exponential. At $x = 0$ the false formula gives $0 \cdot 2^{-1} = 0$, a horizontal tangent, while the figure shows $y = 2^x$ strictly increasing through $(0, 1)$ with slope $\ln 2$. Before differentiating, ask where the variable is: in the base (power rule) or in the exponent (exponential rule). A variable in both, like $x^x$, needs another tool, met later in the course.