Corrected exercises: Differentiation rules (MATH 203)

Rewrite: $\frac{3}{2x^4} = \frac{3}{2}x^{-4}$ and $\frac{x^3}{6} = \frac{1}{6}x^3$. Power rule term by term: $\frac{3}{2} \cdot (-4)x^{-5} = -6x^{-5}$, $\frac{1}{6} \cdot 3x^2 = \frac{1}{2}x^2$, $(5x)' = 5$. So $f'(x) = -\frac{6}{x^5} - \frac{x^2}{2} + 5$. At $x = 2$: $-\frac{6}{32} - 2 + 5 = 3 - \frac{3}{16} = \frac{45}{16} = 2.8125$. The classic slip is $\frac{3}{2x^4} = \frac{3}{2}x^{4}$ or $(2x)^{-4}$: the first forgets that a power in the denominator becomes NEGATIVE, the second raises the $2$ to the power $-4$ as well, and both change every later number.

Rewrite with the laws of exponents: $x^2\sqrt x = x^2 \cdot x^{1/2} = x^{5/2}$ (exponents ADD when powers multiply), and $\frac{4}{\sqrt[3]{x^2}} = 4x^{-2/3}$. Then $g'(x) = \frac{5}{2}x^{3/2} - 4 \cdot \left(-\frac{2}{3}\right)x^{-5/3} = \frac{5}{2}x^{3/2} + \frac{8}{3}x^{-5/3}$. Back to radicals: $x^{3/2} = x\sqrt x$ and $x^{5/3} = x\sqrt[3]{x^2}$, so $g'(x) = \frac{5}{2}x\sqrt x + \frac{8}{3x\sqrt[3]{x^2}}$. At $x = 1$ every power is $1$: $g'(1) = \frac{5}{2} + \frac{8}{3} = \frac{31}{6}$. The exponent arithmetic is the whole question: $-\frac{2}{3} - 1 = -\frac{5}{3}$, since subtracting $1$ moves a negative exponent AWAY from zero.

The denominator is one power of $x$: $x\sqrt x = x^{3/2}$, so divide each term: $h(x) = 2x \cdot x^{-3/2} - 3x^{-3/2} = 2x^{-1/2} - 3x^{-3/2}$. Then $h'(x) = -x^{-3/2} + \frac{9}{2}x^{-5/2}$. The smallest power is $x^{-5/2}$ (the most negative exponent), and factoring it out leaves $x^{-3/2} = x^{-5/2} \cdot x$: $h'(x) = x^{-5/2}\left(-x + \frac{9}{2}\right) = \frac{9 - 2x}{2x^{5/2}} = \frac{9 - 2x}{2x^2\sqrt x}$. So $h'(1) = \frac{7}{2}$. The tangent is horizontal where the numerator is zero, the denominator being positive for $x > 0$: $x = \frac{9}{2}$. Factoring out $x^{-3/2}$ instead, the larger power, leaves $x^{-1}$ inside the bracket and a fraction inside a fraction: always factor the SMALLEST power, as with $x^2 + x^3 = x^2(1 + x)$.

First line: in $\frac{1}{2x^3}$ only $x^3$ is in the denominator as a power; the $2$ is a plain number. So $\frac{1}{2x^3} = \frac{1}{2}x^{-3}$, never $2x^{-3}$, and the derivative is $\frac{1}{2} \cdot (-3)x^{-4} = -\frac{3}{2x^4}$. The student's answer is four times too large: at $x = 1$ it gives $-6$ against the true $-\frac{3}{2}$. Second line: the exponent was increased by $1$ instead of decreased, $-\frac{2}{3} + 1 = \frac{1}{3}$. The rule subtracts $1$: $-\frac{2}{3} - 1 = -\frac{5}{3}$, so $\frac{d}{dx}x^{-2/3} = -\frac{2}{3}x^{-5/3} = -\frac{2}{3x\sqrt[3]{x^2}}$. A quick sanity check: $x^{-2/3}$ decreases for $x > 0$, and $-\frac{2}{3}x^{1/3}$ happens to be negative too, so the sign does not catch this one; the scientific calculator does, with $\frac{f(1.001) - f(0.999)}{0.002}$.

$\frac{e^x}{3} = \frac{1}{3}e^x$ has derivative $\frac{1}{3}e^x$. In $\frac{e^3}{x}$ the numerator $e^3 \approx 20.09$ is a CONSTANT coefficient, so rewrite $\frac{e^3}{x} = e^3x^{-1}$, whose derivative is $-e^3x^{-2}$. Hence $p'(x) = \frac{e^x}{3} - \frac{e^3}{x^2}$, and $p'(3) = \frac{e^3}{3} - \frac{e^3}{9} = \frac{2e^3}{9} \approx 4.4635$. Treating $e^3$ as a function (writing $3e^2$, or using the quotient rule with $(e^3)' = e^3$) are the two ways to lose this mark; a letter that does not contain $x$ is a number.

Product rule with $f(x) = x^{-1} + 2$, $f'(x) = -x^{-2}$, and $g(x) = x^2 - 5$, $g'(x) = 2x$: $F'(x) = -x^{-2}(x^2 - 5) + (x^{-1} + 2)(2x) = -1 + 5x^{-2} + 2 + 4x = 4x + 1 + \frac{5}{x^2}$. The products $x^{-2} \cdot x^2 = 1$ and $x^{-1} \cdot 2x = 2$ are laws of exponents, and they are where the algebra goes wrong. Check by expanding first: $F(x) = x - 5x^{-1} + 2x^2 - 10$, so $F'(x) = 1 + 5x^{-2} + 4x$, the same. $F'(1) = 4 + 1 + 5 = 10$.

Quotient rule with $f(x) = x^2 - 4$, $f'(x) = 2x$, and $g(x) = 2x + 1$, $g'(x) = 2$: $G'(x) = \frac{2x(2x + 1) - (x^2 - 4) \cdot 2}{(2x + 1)^2}$. The minus sign multiplies the WHOLE product: $-(x^2 - 4) \cdot 2 = -2x^2 + 8$. So the numerator is $4x^2 + 2x - 2x^2 + 8 = 2x^2 + 2x + 8$ and $G'(x) = \frac{2x^2 + 2x + 8}{(2x + 1)^2}$, with $G'(0) = \frac{8}{1} = 8$. Writing $-2x^2 - 8$ gives $G'(0) = -8$: one missing pair of parentheses flips the answer. The denominator stays $(2x + 1)^2$, unexpanded.

Quotient rule with top $e^x$ (derivative $e^x$) and bottom $x^2 + 1$ (derivative $2x$): $H'(x) = \frac{e^x(x^2 + 1) - e^x \cdot 2x}{(x^2 + 1)^2}$. Factor $e^x$ out of the numerator: $e^x(x^2 - 2x + 1) = e^x(x - 1)^2$, a perfect square. So $H'(x) = \frac{e^x(x - 1)^2}{(x^2 + 1)^2}$ and $H'(0) = \frac{1 \cdot 1}{1} = 1$. Since $e^x > 0$ and $(x^2 + 1)^2 > 0$, $H'(x) = 0$ only when $(x - 1)^2 = 0$: one horizontal tangent, at $x = 1$, where $H(1) = \frac{e}{2}$. Recognising $x^2 - 2x + 1$ as $(x - 1)^2$ is the step that makes the question readable; the factored form also shows that $H'$ is never negative.

Multiply the top and the bottom by $x$, the common denominator of the small fractions: $K(x) = \frac{x - 2}{x^2 + 3}$, valid for $x \ne 0$ only, since the original expression is undefined at $0$. Quotient rule: $K'(x) = \frac{1 \cdot (x^2 + 3) - (x - 2) \cdot 2x}{(x^2 + 3)^2} = \frac{x^2 + 3 - 2x^2 + 4x}{(x^2 + 3)^2} = \frac{-x^2 + 4x + 3}{(x^2 + 3)^2}$, for $x \ne 0$. At $x = 1$: $\frac{-1 + 4 + 3}{16} = \frac{3}{8}$. Differentiating the complex fraction directly is legal but produces fractions of fractions in every term; clearing it first is the standard gesture, and the condition $x \ne 0$ must travel with the simplified form.

No chain rule yet, so expand the denominator: $(x + 1)^2 = x^2 + 2x + 1$, derivative $2x + 2$. Quotient rule: $q'(x) = \frac{(x^2 + 2x + 1) - x(2x + 2)}{(x + 1)^4} = \frac{1 - x^2}{(x + 1)^4}$, using $(x^2 + 2x + 1)^2 = (x + 1)^4$. Now factor the difference of squares, $1 - x^2 = (1 - x)(1 + x)$, and cancel one factor $x + 1$: $q'(x) = \frac{1 - x}{(x + 1)^3}$. Then $q'(2) = \frac{-1}{27}$. The tangent is horizontal where $1 - x = 0$, at $x = 1$ (the point $\left(1, \frac{1}{4}\right)$). Reading the zeros of $1 - x^2$ BEFORE cancelling gives $x = \pm 1$, but $x = -1$ is not in the domain of $q$: it must be rejected, and saying so is part of the answer.

Rewrite $\frac{4}{x} = 4x^{-1}$: $f'(x) = 1 - 4x^{-2} = 1 - \frac{4}{x^2} = \frac{x^2 - 4}{x^2}$, over the common denominator $x^2$. At $a = 1$: $f(1) = 5$ and $f'(1) = -3$, so the tangent is $y = 5 - 3(x - 1) = -3x + 8$. Check: at $x = 1$ the line gives $5$. The derivative of $\frac{4}{x}$ written as $\frac{0}{1} = 0$, or as $4$, are the two slips that make this part worthless; the rewriting $4x^{-1}$ avoids both.

The tangent slope is $-3 \ne 0$, so the normal slope is $-\frac{1}{-3} = \frac{1}{3}$, and the normal passes through $(1, 5)$: $y = 5 + \frac{1}{3}(x - 1) = \frac{1}{3}x + \frac{14}{3}$. The constant needs a common denominator, $5 - \frac{1}{3} = \frac{15}{3} - \frac{1}{3} = \frac{14}{3}$: this fraction arithmetic is where the question is usually lost, not in the slope. Check: $(-3) \cdot \frac{1}{3} = -1$.

$f'(x) = 0$ when the numerator is zero and the denominator is not: $x^2 - 4 = 0$, $x = \pm 2$, both in the domain. $f(2) = 2 + 2 = 4$ and $f(-2) = -2 - 2 = -4$. The points are $(2, 4)$ and $(-2, -4)$, the bottom of the right branch and the top of the left one on the figure.

Solve the line for $y$: $-4y = 7 - 3x$, so $y = \frac{3}{4}x - \frac{7}{4}$, slope $\frac{3}{4}$. Reading the slope as $3$ (the coefficient of $x$ before solving) is the classic error. Parallel means equal slopes: $1 - \frac{4}{x^2} = \frac{3}{4}$, so $\frac{4}{x^2} = \frac{1}{4}$, $x^2 = 16$, $x = \pm 4$. $f(4) = 4 + 1 = 5$ and $f(-4) = -5$: the points are $(4, 5)$ and $(-4, -5)$. Note also that $f'(x) = 1 - \frac{4}{x^2} < 1$ for every $x$: no tangent is parallel to a line of slope $1$ or more.

Intersect the normal with the curve: $\frac{1}{3}x + \frac{14}{3} = x + \frac{4}{x}$. Multiply by $3x$ (allowed, $x \ne 0$): $x^2 + 14x = 3x^2 + 12$, so $2x^2 - 14x + 12 = 0$, $x^2 - 7x + 6 = 0$, $(x - 1)(x - 6) = 0$. The root $x = 1$ is the point we started from, which is a good check; the new point has $x = 6$ and $y = 6 + \frac{4}{6} = \frac{20}{3}$. On the normal, $\frac{6}{3} + \frac{14}{3} = \frac{20}{3}$: consistent. The solution figure shows the tangent, the normal, and the two points.

Two conditions, two unknowns. The parabola passes through the point of the line at $x = 2$, which is $(2, 2)$: $4 + 2b + c = 2$. Its slope there equals the slope $1$ of the line: $y' = 2x + b$, so $4 + b = 1$ and $b = -3$. Then $c = 2 - 4 + 6 = 4$, and $y = x^2 - 3x + 4$. Check: at $x = 2$, $y = 4 - 6 + 4 = 2$ and $y' = 4 - 3 = 1$. Using only the slope condition leaves $c$ undetermined: tangency is TWO equations, same point and same slope.

Rewrite $\frac{b}{x} = bx^{-1}$, so $f'(x) = a - \frac{b}{x^2}$. The point: $f(2) = 2a + \frac{b}{2} = 8$. The horizontal tangent: $f'(2) = a - \frac{b}{4} = 0$, so $b = 4a$. Substituting: $2a + 2a = 8$, $a = 2$, $b = 8$. The function is $f(x) = 2x + \frac{8}{x}$, and indeed $f(2) = 4 + 4 = 8$ and $f'(2) = 2 - \frac{8}{4} = 0$. Here $b$ is a constant in a numerator: $\left(\frac{b}{x}\right)' = -\frac{b}{x^2}$, not $\frac{1}{x}$ or $0$.

The slopes are $3x^2 - 1$ and $2x$, both at the SAME $x$ (the question compares the two curves above one value of $x$). Parallel tangents: $3x^2 - 1 = 2x$, $3x^2 - 2x - 1 = 0$, $(3x + 1)(x - 1) = 0$. So $x = -\frac{1}{3}$ or $x = 1$. At $x = 1$ both slopes are $2$; at $x = -\frac{1}{3}$ both are $-\frac{2}{3}$. Forgetting to move $2x$ to one side and factoring $3x^2 - 1$ alone is the usual way to lose this part.

$(e^x)' = e^x$, so solve $e^x = 3$: $x = \ln 3 \approx 1.0986$, and the point is $(\ln 3, 3)$, since $e^{\ln 3} = 3$ exactly. The tangent is $y = 3 + 3(x - \ln 3) = 3x + 3 - 3\ln 3$, with $y$-intercept $3 - 3\ln 3 \approx -0.2958$. Keep $\ln 3$ exact until the last line: rounding $x$ to $1.1$ first gives an intercept of $-0.3$, wrong at the second decimal.

Solve the line for $y$: $y = -\frac{1}{12}x + \frac{5}{12}$, slope $-\frac{1}{12}$. A normal with slope $-\frac{1}{12}$ is perpendicular to a tangent of slope $12$, since $12 \cdot \left(-\frac{1}{12}\right) = -1$. So solve $y' = 6x^2 - 6x = 12$: $x^2 - x - 2 = 0$, $(x - 2)(x + 1) = 0$, $x = 2$ or $x = -1$. The points are $(2, 16 - 12) = (2, 4)$ and $(-1, -2 - 3) = (-1, -5)$. Setting $y' = -\frac{1}{12}$ answers a different question, the one where the TANGENT is parallel to the line.