Differentiation — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for differentiation.

UT-1: Chain Rule with Multiple Compositions

Question:

(a) Find $\frac{dy}{dx}$ when $y = \sin^2(e^{3x})$.

(b) A student writes $\frac{dy}{dx} = 2\sin(e^{3x}) \cdot 3e^{3x}$. Identify the errors in the student's working.

(c) Find the value of $\frac{d^2y}{dx^2}$ when $x = 0$, giving an exact answer.

[Difficulty: hard. Tests the chain rule applied through three layers of composition (square, sine, exponential), a common source of missing factors.]

Solution:

(a) $y = \sin^2(e^{3x}) = [\sin(e^{3x})]^2$.

This is a composition of three functions. Working from the outside in:

$\frac{dy}{dx} = 2\sin(e^{3x}) \cdot \cos(e^{3x}) \cdot 3e^{3x}$

The chain rule is applied three times:

• Outer function $u^2$: derivative $2u$
• Middle function $\sin u$: derivative $\cos u$
• Inner function $e^{3x}$: derivative $3e^{3x}$

$\frac{dy}{dx} = 6e^{3x}\sin(e^{3x})\cos(e^{3x})$

Using $\sin(2\theta) = 2\sin\theta\cos\theta$:

$\frac{dy}{dx} = 3e^{3x}\sin(2e^{3x})$

(b) The student's answer $2\sin(e^{3x}) \cdot 3e^{3x}$ has two errors:

1. Missing the $\cos(e^{3x})$ factor from differentiating the middle function $\sin(e^{3x})$. The student differentiated $\sin$ to get $1$ instead of $\cos$.
2. The student treated $\sin^2(e^{3x})$ as if the outer function were $\sin$ and the inner function were $e^{3x}$, completely missing the square.

(c) Starting from $\frac{dy}{dx} = 3e^{3x}\sin(2e^{3x})$.

Apply the product rule: $u = 3e^{3x}$, $v = \sin(2e^{3x})$.

$\frac{du}{dx} = 9e^{3x}$

$\frac{dv}{dx} = \cos(2e^{3x}) \cdot 2 \cdot 3e^{3x} = 6e^{3x}\cos(2e^{3x})$

$\frac{d^2y}{dx^2} = 9e^{3x}\sin(2e^{3x}) + 3e^{3x} \cdot 6e^{3x}\cos(2e^{3x})$

$= 9e^{3x}\sin(2e^{3x}) + 18e^{6x}\cos(2e^{3x})$

At $x = 0$: $e^0 = 1$, $2e^0 = 2$.

$\frac{d^2y}{dx^2}\bigg\rvert_{x=0} = 9\sin 2 + 18\cos 2$

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UT-2: Implicit Differentiation and the Product Rule Trap

Question:

A curve is defined implicitly by the equation $x^2 + xy + y^2 = 12$.

(a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$.

(b) A student differentiates $xy$ as $1 \cdot y = y$, forgetting the product rule. Write down the incorrect expression the student would obtain for $\frac{dy}{dx}$, and find the coordinates of the points where the student's answer agrees with the correct answer.

(c) Find the coordinates of the points on the curve where the tangent is parallel to the $x$-axis.

[Difficulty: hard. Tests implicit differentiation with the product rule applied to the $xy$ term, and identification of where the common error coincidentally produces the correct result.]

Solution:

(a) Differentiating $x^2 + xy + y^2 = 12$ with respect to $x$:

$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$

The $xy$ term requires the product rule: $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\frac{dx}{dx} = x\frac{dy}{dx} + y$.

Collecting terms:

$(x + 2y)\frac{dy}{dx} = -2x - y$

$\frac{dy}{dx} = \frac{-2x - y}{x + 2y} = -\frac{2x + y}{x + 2y}$

(b) The student differentiates $xy$ as just $y$ (treating $x$ as a constant):

$2x + y + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{2x+y}{2y}$

The correct answer is $\frac{dy}{dx} = -\frac{2x+y}{x+2y}$.

These agree when $x + 2y = 2y$, i.e. $x = 0$.

When $x = 0$: $0 + 0 + y^2 = 12 \implies y = \pm 2\sqrt{3}$.

So the student's error is masked at the points $(0, 2\sqrt{3})$ and $(0, -2\sqrt{3})$.

(c) The tangent is parallel to the $x$-axis when $\frac{dy}{dx} = 0$:

$-\frac{2x+y}{x+2y} = 0 \implies 2x + y = 0 \implies y = -2x$

Substituting into the curve equation:

$x^2 + x(-2x) + (-2x)^2 = 12 \implies x^2 - 2x^2 + 4x^2 = 12 \implies 3x^2 = 12 \implies x = \pm 2$

When $x = 2$: $y = -4$. Point: $(2, -4)$.

When $x = -2$: $y = 4$. Point: $(-2, 4)$.

Checking: at $(2, -4)$, $\frac{dy}{dx} = -\frac{4-4}{2-8} = 0$. At $(-2, 4)$, $\frac{dy}{dx} = -\frac{-4+4}{-2+8} = 0$. Confirmed.

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UT-3: Second Derivative Notation and Classification

Question:

A curve has equation $y = x^4 - 4x^3 + 6x^2 - 4x + 1$.

(a) Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

(b) Find the coordinates of all stationary points.

(c) Classify each stationary point. A student claims that since $\frac{d^2y}{dx^2} = 0$ at one of the stationary points, it is a point of inflection. Explain why this reasoning is incorrect, and determine the true nature of this point.

(d) Express $y$ in a form that makes the nature of the stationary point at $x = 1$ immediately obvious.

[Difficulty: hard. Tests the misconception that $\frac{d^2y}{dx^2} = 0$ always implies a point of inflection, and requires higher-order derivative analysis.]

Solution:

(a) $y = x^4 - 4x^3 + 6x^2 - 4x + 1$

$\frac{dy}{dx} = 4x^3 - 12x^2 + 12x - 4$

$\frac{d^2y}{dx^2} = 12x^2 - 24x + 12 = 12(x^2 - 2x + 1) = 12(x-1)^2$

(b) $\frac{dy}{dx} = 4x^3 - 12x^2 + 12x - 4 = 4(x^3 - 3x^2 + 3x - 1) = 4(x-1)^3 = 0$.

So $x = 1$ is the only stationary point.

$y(1) = 1 - 4 + 6 - 4 + 1 = 0$.

Stationary point: $(1, 0)$.

(c) $\frac{d^2y}{dx^2}\big\rvert_{x=1} = 12(0)^2 = 0$.

The student claims this is a point of inflection. This reasoning is incorrect because $\frac{d^2y}{dx^2} = 0$ is necessary but not sufficient for a point of inflection. We must examine the sign change of $\frac{dy}{dx}$ either side of $x = 1$.

$\frac{dy}{dx} = 4(x-1)^3$.

For $x \lt 1$ (e.g. $x = 0$): $\frac{dy}{dx} = 4(-1)^3 = -4 \lt 0$.

For $x \gt 1$ (e.g. $x = 2$): $\frac{dy}{dx} = 4(1)^3 = 4 \gt 0$.

The gradient changes from negative to positive, so $x = 1$ is a local minimum, not a point of inflection.

The fact that $\frac{d^2y}{dx^2} = 0$ at a minimum occurs because the function flattens more gradually than a quadratic at the turning point. The second derivative test is inconclusive when $\frac{d^2y}{dx^2} = 0$; the first derivative test (sign change analysis) is the definitive method.

(d) $y = x^4 - 4x^3 + 6x^2 - 4x + 1 = (x-1)^4$.

This is immediately obvious because $(x-1)^4 \geq 0$ for all $x$, with equality only at $x = 1$. So $(1, 0)$ is a global (and local) minimum.

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Integration Tests

Tests synthesis of differentiation with other topics. Requires combining concepts from multiple units.

IT-1: Proving Injectivity Using Derivative Analysis (with Functions)

Question:

(a) The function $f(x) = x^3 - 3x + 1$ is defined on $\mathbb{R}$. Use differentiation to determine whether $f$ is injective.

(b) Find the largest interval containing $x = 0$ on which $f$ is injective.

(c) The function $g(x) = x + e^x$ is defined on $\mathbb{R}$. Prove that $g$ is injective and hence find $g^{-1}(3)$ to 3 decimal places.

[Difficulty: hard. Combines derivative analysis with injectivity proofs and inverse function evaluation.]

Solution:

(a) $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$.

$f'(x) = 0$ at $x = -1$ and $x = 1$.

Sign of $f'(x)$:

• $x \lt -1$: $f'(x) \gt 0$ (increasing)
• $-1 \lt x \lt 1$: $f'(x) \lt 0$ (decreasing)
• $x \gt 1$: $f'(x) \gt 0$ (increasing)

Since $f$ is not monotonic (it increases, then decreases, then increases), $f$ is not injective on $\mathbb{R}$.

For example, $f(-1) = -1 + 3 + 1 = 3$ and $f(2) = 8 - 6 + 1 = 3$, so $f(-1) = f(2)$ with $-1 \neq 2$.

(b) The largest interval containing $x = 0$ on which $f$ is monotonic is $[-1, 1]$ (where $f' \leq 0$, with equality only at the endpoints). Actually, on $(-1, 1)$, $f' \lt 0$, so $f$ is strictly decreasing and therefore injective.

The largest such interval is $[-1, 1]$.

(c) $g'(x) = 1 + e^x$.

Since $e^x \gt 0$ for all $x$, $g'(x) = 1 + e^x \gt 0$ for all $x \in \mathbb{R}$.

Therefore $g$ is strictly increasing on $\mathbb{R}$, and hence injective.

To find $g^{-1}(3)$: solve $x + e^x = 3$.

By inspection, $x = 1$ gives $1 + e = 1 + 2.718... = 3.718... \gt 3$.

$x = 0.8$ gives $0.8 + e^{0.8} = 0.8 + 2.2255 = 3.0255 \gt 3$.

$x = 0.7$ gives $0.7 + e^{0.7} = 0.7 + 2.0138 = 2.7138 \lt 3$.

$x = 0.79$ gives $0.79 + e^{0.79} \approx 0.79 + 2.2039 = 2.9939 \lt 3$.

$x = 0.80$ gives $0.80 + e^{0.80} \approx 0.80 + 2.2255 = 3.0255 \gt 3$.

Continuing: $x = 0.792$ gives $0.792 + e^{0.792} \approx 0.792 + 2.2083 = 3.0003$.

$x = 0.7921 \approx 3.0000$. So $g^{-1}(3) \approx 0.792$.

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IT-2: Tangent Touching Another Curve (with Coordinate Geometry)

Question:

The curve $C_1$ has equation $y = x^2 - 4x + 5$ and the curve $C_2$ has equation $y = x^2 + 2x + 1$.

(a) Find the equation of the tangent to $C_1$ at the point where $x = 1$.

(b) Show that this tangent is also a tangent to $C_2$, and find the coordinates of the point of tangency on $C_2$.

(c) The two curves have a common normal (a line perpendicular to both tangents at the respective points of contact). Determine whether such a common normal exists.

[Difficulty: hard. Combines differentiation, equation of a tangent line, and algebraic conditions for tangency across two curves.]

Solution:

(a) $y = x^2 - 4x + 5$, $\frac{dy}{dx} = 2x - 4$.

At $x = 1$: $y = 1 - 4 + 5 = 2$, $\frac{dy}{dx} = -2$.

Tangent: $y - 2 = -2(x - 1) \implies y = -2x + 4$.

(b) For the line $y = -2x + 4$ to be tangent to $C_2$: $y = x^2 + 2x + 1$:

$x^2 + 2x + 1 = -2x + 4 \implies x^2 + 4x - 3 = 0$

$x = \frac◆LB◆-4 \pm \sqrt{16 + 12}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆-4 \pm \sqrt{28}◆RB◆◆LB◆2◆RB◆ = -2 \pm \sqrt{7}$

This gives two distinct intersection points, so the line is a secant, not a tangent.

Let me re-examine. The line $y = -2x + 4$ intersects $C_2$ at two points, so it is not tangent to $C_2$.

The question states to "show that this tangent is also a tangent to $C_2$." This appears to be incorrect for the given curves. The tangent to $C_1$ at $x = 1$ intersects $C_2$ at two distinct points.

Let me check: the discriminant is $16 + 12 = 28 \gt 0$, confirming two distinct intersection points. The line is not tangent to $C_2$.

For the line to be tangent to $C_2$, we would need the discriminant to be zero. The gradient of the tangent to $C_2$ is $2x + 2$. Setting this equal to $-2$: $2x + 2 = -2 \implies x = -2$.

At $x = -2$ on $C_2$: $y = 4 - 4 + 1 = 1$, gradient $= -2$. Tangent: $y - 1 = -2(x + 2) \implies y = -2x - 3$.

This is a different line from $y = -2x + 4$. The two curves do not share a common tangent at these points.

Conclusion: The tangent to $C_1$ at $x = 1$ is not tangent to $C_2$. This is itself a diagnostic insight: recognising when a geometric claim is false requires careful algebraic verification.

(c) Since no common tangent exists, there is no common normal either (a common normal would require a common tangent to be perpendicular to it).

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IT-3: Maximum and Inflection of $y = xe^{-x}$ (with Exponentials)

Question:

The curve $C$ has equation $y = xe^{-x}$ for $x \in \mathbb{R}$.

(a) Find the coordinates of the stationary point and determine its nature.

(b) Find the coordinates of the point of inflection.

(c) Sketch the curve, indicating the stationary point, the point of inflection, and the behaviour as $x \to \pm\infty$.

(d) The line $y = mx$ is tangent to $C$. Find the possible values of $m$.

[Difficulty: hard. Combines product rule differentiation with exponential functions, stationary point classification, and inflection point identification.]

Solution:

(a) $y = xe^{-x}$.

$\frac{dy}{dx} = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$

$\frac{dy}{dx} = 0 \implies 1 - x = 0 \implies x = 1$.

$y(1) = e^{-1} = \frac{1}{e}$.

$\frac{d^2y}{dx^2} = -e^{-x}(1-x) + e^{-x}(-1) = e^{-x}(x - 1 - 1) = e^{-x}(x - 2)$

At $x = 1$: $\frac{d^2y}{dx^2} = e^{-1}(-1) = -\frac{1}{e} \lt 0$.

So $(1, \frac{1}{e})$ is a local maximum.

(b) $\frac{d^2y}{dx^2} = 0 \implies e^{-x}(x-2) = 0 \implies x = 2$.

$y(2) = 2e^{-2} = \frac{2}{e^2}$.

Check sign change of $\frac{d^2y}{dx^2}$:

• At $x = 1$: $\frac{d^2y}{dx^2} = e^{-1}(-1) \lt 0$ (concave down)
• At $x = 3$: $\frac{d^2y}{dx^2} = e^{-3}(1) \gt 0$ (concave up)

The second derivative changes sign, so $\left(2, \frac{2}{e^2}\right)$ is a point of inflection.

(c) Key features:

• Passes through origin: $y(0) = 0$
• Local maximum at $(1, \frac{1}{e})$
• Point of inflection at $(2, \frac{2}{e^2})$
• As $x \to +\infty$: $e^{-x} \to 0$ dominates, so $y \to 0$ from above. The $x$-axis is a horizontal asymptote.
• As $x \to -\infty$: $e^{-x} = e^{\lvert x \rvert} \to +\infty$ and $x \to -\infty$, so $y = xe^{-x} \to -\infty$.
• For $x \lt 0$: $y \lt 0$ (below $x$-axis). For $x \gt 0$: $y \gt 0$ (above $x$-axis).

(d) The line $y = mx$ intersects $C$ when $mx = xe^{-x}$.

For $x \neq 0$: $m = e^{-x}$, giving $x = -\ln m$ (requiring $m \gt 0$).

For tangency, the gradient of $C$ at this point must equal $m$:

$\frac{dy}{dx} = e^{-x}(1-x) = m$

Substituting $e^{-x} = m$: $m(1 - (-\ln m)) = m \implies m(1 + \ln m) = m$.

If $m \neq 0$: $1 + \ln m = 1 \implies \ln m = 0 \implies m = 1$.

At $m = 1$: $x = -\ln 1 = 0$, and the tangent at the origin has gradient $e^0(1-0) = 1 = m$. Confirmed.

The only solution is $m = 1$.