Implicit and Parametric Differentiation — Tangents and Normals

The curve $x^3+y^3=6xy$ passes through the point $(3,3)$. Find $\dfrac{dy}{dx}$ at this point.

• $-1$, from $3x^2+3y^2\dfrac{dy}{dx}=6y+6x\dfrac{dy}{dx}$ (product rule on the right side), giving $27+27\dfrac{dy}{dx}=18+18\dfrac{dy}{dx}$ at $(3,3)$, so $9\dfrac{dy}{dx}=-9$

  Correct. Both sides need care: the left needs the chain rule on y³ (3y²·dy/dx), and the right needs the PRODUCT rule on 6xy (6y + 6x·dy/dx, not just 6y) — a term this lesson's own trap taxonomy names directly as one of two rules most often only half-applied.

• B$-\dfrac{1}{3}$, from treating $6xy$ as differentiating to $6y$ only

  This drops the product rule's second term entirely, treating x as though it were a constant multiplier rather than a variable being differentiated too — the same gap this lesson's own trap taxonomy documents for a different mixed term (3x²y → 6x dy/dx, from Oct 2021 Q1).

• C$2$, from writing the left side as $3x^2+3y^2$ with no $\dfrac{dy}{dx}$ attached to the $y^3$ term

  y³ is a function of y, and y is a function of x — differentiating it with respect to x needs the chain rule, which multiplies by dy/dx. Leaving that factor off treats y as though it were a fixed constant, not the unknown function the whole equation is built around.

• D$1$, from a sign slip collecting the $\dfrac{dy}{dx}$ terms onto one side

  The correct rearrangement is 3y²·dy/dx − 6x·dy/dx = 6y − 3x², i.e. 27·dy/dx − 18·dy/dx = 18 − 27; losing a sign on either side of that move produces a plausible-looking but wrong result like this one — worth checking by re-substituting dy/dx = −1 back into the very first differentiated equation, which balances: 27 + 27(−1) = 0 = 18 + 18(−1).

Traps tested: Product rule term dropped differentiating a mixed xy term · Chain rule factor omitted entirely · Sign error collecting dy dx terms

Curve $C$: $x^2+y^2+6x-8y=0$. $Q$ is the point where $C$ crosses the negative $x$-axis, other than the origin. A candidate correctly finds $Q=(-6,0)$, correctly derives $\dfrac{dy}{dx}=\dfrac{-3-x}{y-4}$, but then evaluates it at the origin instead of at $Q$. What gradient do they get, and is it the correct gradient for the tangent at $Q$?

• They get $\dfrac{3}{4}$ — but the actual gradient at $Q$ is $-\dfrac{3}{4}$, so their answer is wrong despite every earlier step being right

  Correct. At the origin: (−3−0)/(0−4) = (−3)/(−4) = 3/4. At Q: (−3−(−6))/(0−4) = 3/(−4) = −3/4 — the two gradients aren't even the same sign. This is exactly the documented real trap applied to a fresh curve: the coordinates and the differentiation can both be entirely correct, and the tangent still comes out wrong from one substitution.

• BThey get $\dfrac{3}{4}$, and that is also the correct gradient at $Q$, since both points lie on the same curve

  Lying on the same curve doesn't make two different points share a gradient — a curve's steepness generally changes from point to point, which is the entire reason dy/dx for an implicit curve comes out as an expression in x AND y, rather than a single fixed number.

• CThey get $-\dfrac{3}{4}$, the same as the correct value, since the origin and $Q$ are symmetric on the curve

  There is no such symmetry here to rely on, and the arithmetic doesn't support it either: substituting (0,0) into (−3−x)/(y−4) gives (−3)/(−4) = 3/4, the positive value, not −3/4.

• DThe gradient cannot be found at either point, since $C$ passes through the origin twice

  C passes through the origin once, like any other point on it — 'crosses the x-axis at two points, one of which happens to be the origin' is not the same claim as 'passes through the origin twice.' Nothing about the origin lying on C prevents the gradient formula from being evaluated there; the problem is only that the ORIGIN is the wrong point for this question.

Traps tested: Gradient assumed constant across a curve · Gradient values assumed equal without checking · Curve passing through origin misread as a double point

An implicit equation contains the isolated term $5x^2y$. Differentiated correctly with respect to $x$, what does this term become?

• $10xy + 5x^2\dfrac{dy}{dx}$

  Correct — the product rule applied to u = 5x², v = y: u·dv/dx + v·du/dx = 5x²·dy/dx + y(10x) = 10xy + 5x²·dy/dx.

• B$10x\dfrac{dy}{dx}$

  This differentiates the x²-part only (to 10x) and appends a bare dy/dx, which is exactly the documented real slip pattern for a term of this shape: 'differentiating 3x²y → 6x dy/dx' (Oct 2021, Q1) — the product rule's other term, 10xy, is missing entirely, and the surviving term should be 5x²·dy/dx, not 10x·dy/dx.

• C$10xy$

  This differentiates only the x² part of the product (correctly, to 10x, times y) and drops the OTHER product-rule term entirely — the one that comes from differentiating y itself, which needs the chain rule and contributes 5x²·dy/dx.

• D$5x^2\cdot 2y\dfrac{dy}{dx} = 10x^2y\dfrac{dy}{dx}$

  This treats 5x²y as though it were 5x²·y² (applying the squared-y chain rule to a term where y only appears to the first power) — the correct chain-rule contribution for a single power of y is dy/dx itself, not 2y·dy/dx, and the product rule's other term is still missing besides.

Traps tested: Product rule term dropped differentiating a mixed xy term · Product rule half of term omitted · Chain rule applied as if y were squared

An implicit equation contains the isolated term $3y^2$. Differentiated correctly with respect to $x$, what does this term become?

• $6y\dfrac{dy}{dx}$

  Correct — the chain rule: d/dx(3y²) = 3 × 2y × dy/dx = 6y·dy/dx, the coefficient 3 carried straight through unchanged.

• B$6\dfrac{dy}{dx}$

  This is exactly the documented real slip for a term of this shape: 'differentiating 4y² → 8 dy/dx' (Oct 2021, Q1), scaled to a different coefficient — the leftover factor of y from the chain rule (d/dy(y²) = 2y, not 1) has been dropped, keeping only the dy/dx half of what the chain rule actually produces.

• C$6y$

  This differentiates y² as though y were simply x in disguise — correct for a term like 3x², but y is not the variable being differentiated WITH RESPECT TO; it is itself a function of that variable, which is exactly why the chain rule attaches a dy/dx factor here and would not for 3x².

• D$0$, since $y$ is not a known function of $x$ and cannot be differentiated

  Not knowing y's explicit formula doesn't make it undifferentiable — implicit differentiation exists specifically for this situation: y is treated as SOME function of x, whatever that function turns out to be, and the chain rule works without ever needing to know its formula.

Traps tested: Chain rule y factor dropped differentiating a y power term · Y treated as the independent variable · Unknown formula confused with undifferentiable

An implicit equation contains the isolated term $7x^3+4$. Differentiated correctly with respect to $x$, what does this become?

• $21x^2$

  Correct, and worth noticing what is absent: no dy/dx appears anywhere, because there is no y anywhere in this term — 7x³ differentiates by the ordinary power rule, and the constant 4 differentiates to 0.

• B$21x^2+4\dfrac{dy}{dx}$

  This is exactly the documented real slip WMA14-verified-facts.md itself calls out as surprising: 'differentiating 4x² + 8 → 8x + 8 dy/dx' (Oct 2021, Q1) — a dy/dx reflexively attached to a constant term that contains no y at all. The 'y-terms pick up dy/dx' rule only applies where there is a y to begin with.

• C$21x^2+4$

  The constant 4 does not survive differentiation unchanged — the derivative of ANY constant is 0, whether or not the term next to it contains x. This is a different, and equally real, slip from the one this question is really testing, but still wrong.

• D$21x^2\dfrac{dy}{dx}$

  There is no reason for a dy/dx to appear anywhere in this differentiation, and multiplying the WHOLE result by one compounds the same reflexive-attachment error in a different place — the constant 4 has also vanished from this answer entirely, a second, separate error on top of the first.

Traps tested: Dy dx added reflexively to a term with no y at all · Constant term carried through differentiation unchanged

A curve is given parametrically by $x=t^2+1$, $y=2t^3$. Find an equation of the tangent to the curve at $t=2$.

• $6x-y-14=0$, from $\dfrac{dx}{dt}=2t$, $\dfrac{dy}{dt}=6t^2$, so $\dfrac{dy}{dx}=3t$; at $t=2$ the point is $(5,16)$ and the gradient is $6$

  Correct. dy/dx = 6t²/2t = 3t (cancelling a common factor of 2t, valid for t ≠ 0); at t = 2, gradient = 6, point (2² + 1, 2×2³) = (5,16); tangent y − 16 = 6(x−5) ⇒ y = 6x − 14 ⇒ 6x − y − 14 = 0.

• B$3x-y+1=0$, from $\dfrac{dy}{dx}=\dfrac{dx/dt}{dy/dt}=\dfrac{2t}{6t^2}=\dfrac{1}{3t}$, giving gradient $\dfrac{1}{6}$ at $t=2$

  This forms the RECIPROCAL of dy/dx — dividing dx/dt by dy/dt instead of the other way round. This lesson's own mechanism derives exactly why the correct order is dy/dt ÷ dx/dt: it comes from rearranging dy/dt = dy/dx × dx/dt, which fixes which rate divides which.

• C$6x-y-14=0$ is close, but the point should be $(5,12)$, since $y=2t^3$ at $t=2$ gives $2\times2\times3=12$

  2t³ means 2 × t³, i.e. 2 × (2)³ = 2 × 8 = 16, not 2 × 2 × 3 — the exponent applies to t alone before the coefficient multiplies in, and treating t³ as t × 3 mixes up a power with a product.

• D$3x-y+1=0$, from correctly finding gradient $3t=6$ at $t=2$ but using the point $(4,16)$

  x = t² + 1 at t = 2 is 4 + 1 = 5, not 4 — the +1 is a genuine part of the coordinate and has to be added after squaring t, not dropped.

Traps tested: Parametric derivative ratio inverted · Exponent mistaken for a coefficient multiplication · Constant term dropped evaluating a parametric coordinate

A curve is given parametrically by $x=t^3$, $y=t^2$. Correctly differentiated, $\dfrac{dy}{dx}=\dfrac{2}{3t}$. At $t=2$, a candidate 'simplifies' this — the same style of reciprocal slip documented on a real question of this type — to $\dfrac{3t}{2}$, then substitutes. What gradient do they get, and what is the actual correct gradient at $t=2$?

• They get $3$; the correct gradient is $\dfrac{1}{3}$

  Correct. 2/(3t) at t = 2 is 2/6 = 1/3; inverting it to 3t/2 first and then substituting t = 2 gives 6/2 = 3 — a completely different number, from a fraction that only needed a value substituted into it, not inverted first. Same category of error as the real, quoted trap on this topic — 'simplifying' a compound fraction by flipping it — applied to different numbers.

• BThey get $\dfrac{1}{3}$, the correct value — inverting the fraction and then substituting gives the same final answer either way

  It doesn't: 2/(3t) at t = 2 is 1/3, but 3t/2 at t = 2 is 3 — inverting a fraction changes its value at every t except where the fraction equals exactly 1 or −1, which this one does not at t = 2.

• CThey get $\dfrac{1}{3}$; the correct gradient is actually $3$, since $\dfrac{dy}{dx}$ should be $\dfrac{dx/dt}{dy/dt}$

  This inverts which derivative goes on top — dy/dx is dy/dt over dx/dt, not the other way round, which this lesson's own mechanism derives directly from dy/dt = dy/dx × dx/dt. The value 1/3 given here is (coincidentally) the genuinely correct gradient, reached by the wrong reasoning about which fraction to invert.

• DBoth $3$ and $\dfrac{1}{3}$ are acceptable, since a gradient can be expressed as a fraction or its reciprocal

  A gradient is one specific number, not a value interchangeable with its own reciprocal — 3 and 1/3 describe two visibly different steepnesses (one much steeper than the other), and only one of them is what dy/dx actually evaluates to at this point.

Traps tested: Fraction value assumed unchanged by inversion · Parametric derivative ratio inverted · Gradient and its reciprocal treated as interchangeable

An implicit equation contains the isolated term $ye^{2x}$. Differentiated correctly with respect to $x$, what does this term become?

• $2ye^{2x} + e^{2x}\dfrac{dy}{dx}$

  Correct — the product rule applied to $u=e^{2x}$ (the x-side factor) and $v=y$ (the y-side factor): $u\dfrac{dv}{dx}+v\dfrac{du}{dx} = e^{2x}\dfrac{dy}{dx} + y\left(2e^{2x}\right)$. Every other product-rule term in this lesson pairs the y-factor's chain rule with an x-side factor that is a bare power of x (x, x², 5x², 6x), where $\dfrac{du}{dx}=1$ is, in this lesson's own words, 'the easy half.' Here the x-side factor is $e^{2x}$ itself, which is a composite function and needs its own chain-rule step: $\dfrac{d}{dx}(e^{2x}) = 2e^{2x}$, not just $e^{2x}$ — the x-side derivative is no longer the trivial half.

• B$ye^{2x} + e^{2x}\dfrac{dy}{dx}$

  This gets the product-rule STRUCTURE right — one term carrying dy/dx, one not — but drops the chain-rule coefficient differentiating the x-side factor: $\dfrac{d}{dx}(e^{2x})$ is $2e^{2x}$, not $e^{2x}$, because $e^{2x}$ has its own inner function ($2x$) and needs its own chain-rule multiplier, the same way $\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}$ needs its factor of 2. Treating the x-side factor as though differentiating it were as trivial as differentiating x itself — where the factor genuinely is 1 — is exactly where this term differs from every earlier product-rule example in this lesson.

• C$2ye^{2x}$

  This differentiates the x-side factor correctly (to $2e^{2x}$, times y) and drops the OTHER product-rule term entirely — the one that comes from differentiating y itself, which needs the chain rule and contributes $e^{2x}\dfrac{dy}{dx}$. The same gap this lesson's own trap taxonomy documents for a different mixed term ($3x^2y \to 6x\,dy/dx$, from Oct 2021 Q1), here applied to a term whose x-side factor is transcendental rather than a polynomial power of x.

• D$2ye^{2x} + e^{2x}$

  This keeps both product-rule terms but treats the y-factor as though it were simply x in disguise, differentiating it to 1 instead of $\dfrac{dy}{dx}$ — the chain-rule factor on the y-side, which is not the variable being differentiated with respect to, has been left off entirely.

Traps tested: Chain rule coefficient dropped differentiating an exponential x factor · Product rule term dropped differentiating a mixed xy term · Chain rule factor omitted entirely