Partial derivatives

At a Glance

• Frequency: 8 sub-parts across 6 of 13 years (2013, 2017, 2018, 2019, 2021, 2025)
• Priority tier: T2
• Marks (count): 10 (1), 12 (2), 15 (4), 15 (1)
• Average solve time: ~13 min
• Difficulty mix: medium 5, hard 3
• Section: A | Dominant type: computation

Why This Chapter Matters

Partial-derivatives questions appear in 6 of the last 13 years at 10–15 marks, always in Section A (the compulsory half of the paper). Four distinct archetypes rotate unpredictably: computing mixed second partials at a singular origin where the standard formulas break down; applying Euler’s theorem to disguised homogeneous functions; checking existence of partials for piecewise-defined functions; and using the total differential to approximate values. The mixed-partials-at-origin type is the trickiest — it appears in 2013, 2017, 2021, and 2025 and demands the limit-definition approach. The Euler homogeneous type (2019, 2025) requires recognising which composite expression is homogeneous and extracting the degree. The other two archetypes are mechanical once you know which branch applies.

Minimum Theory

Partial derivatives at singular points. The standard quotient-rule formulas for $f_x$ and $f_y$ are valid only where $f$ is smooth. At an isolated singular point (typically the origin with a $0/0$ form), return to the limit definition: $f_x(0,0)=\lim_{h\to0}[f(h,0)-f(0,0)]/h$. To compute $f_{xy}(0,0)$, first find $f_x(0,y)$ for general $y$, then differentiate that in $y$ at $y=0$.

Schwarz–Clairaut theorem. If $f_{xy}$ and $f_{yx}$ are both continuous at a point then $f_{xy}=f_{yx}$ there. Piecewise functions with a singularity at the origin are designed so that Clairaut’s hypotheses fail, forcing $f_{xy}(0,0)\ne f_{yx}(0,0)$.

Euler’s theorem on homogeneous functions. $\phi(x,y)$ is homogeneous of degree $n$ iff $\phi(tx,ty)=t^n\phi$. First-order: $x\phi_x+y\phi_y=n\phi$. Second-order: $x^2\phi_{xx}+2xy\phi_{xy}+y^2\phi_{yy}=n(n-1)\phi$.

When $f=\psi(u)$ is homogeneous. If $v=\psi(u)$ is homogeneous of degree $n$ but $u$ is not, rewrite Euler’s identity via chain rule: $xu_x+yu_y=nv/\psi'(u)=:G(u)$. The second-order identity becomes $x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=G(u)[G'(u)-1]$.

Total differential. $f(x_0+\Delta x,y_0+\Delta y)\approx f(x_0,y_0)+f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y$. Choose $(x_0,y_0)$ to make $f(x_0,y_0)$ a clean number.

Question Archetypes

Four patterns cover every partial-derivatives question in the corpus.

mixed-partials-at-origin (4 question(s); 2013, 2017, 2021, 2025)

Recognition Cues

• Piecewise $f(0,0)=0$ with formula involving $xy/(\text{something})$ or $|x^2-y^2|$; “find $f_{xy}(0,0)$ and $f_{yx}(0,0)$.”
• “Show they are equal” or “show they are unequal.”
• Continuity of $f_{xy}$ and $f_{yx}$ at the origin is often part of the required answer.

Solution Template

1. First partials away from origin — quotient/chain rule on the formula.
2. First partials at origin — limit definition: $f_x(0,0)=\lim_{h\to0}f(h,0)/h$ (assuming $f(0,0)=0$); usually both are $0$.
3. $f_x(0,y)$ — evaluate $\lim_{h\to0}[f(h,y)-f(0,y)]/h$ for general $y$ (holding $y$ fixed). This is a one-variable limit giving a function of $y$.
4. $f_{xy}(0,0)$ — differentiate $f_x(0,y)$ at $y=0$: $\lim_{k\to0}[f_x(0,k)-0]/k$.
5. $f_y(x,0)$ and $f_{yx}(0,0)$ — same procedure with $x$ and $y$ swapped.
6. Continuity — approach along $y=0$ and $x=0$; if limits differ, $f_{xy}$ is discontinuous at the origin.

Worked Example(s)

2013 Paper 1, 2013-P1-Q3b (15 marks)

$f(x,y)=xy^3/(x+y^2)$ for $(x,y)\ne(0,0)$, $f(0,0)=0$. Compute $f_{xy}(0,0)$ and $f_{yx}(0,0)$. Discuss continuity of $f_{xy}$ and $f_{yx}$ at $(0,0)$.

Partials away from origin: $f_x=y^5/(x+y^2)^2$, $f_y=xy^2(3x+y^2)/(x+y^2)^2$.

At origin: $f_x(0,0)=f_y(0,0)=0$.

$f_x(0,y)$: $\lim_{h\to0}[f(h,y)-0]/h=\lim_{h\to0}y^3/(h+y^2)$. For $y\ne0$: $f_x(0,y)=y^3/y^2=y$. So: $f_{xy}(0,0)=\lim_{k\to0}\frac{k-0}{k}=\mathbf{1}.$

$f_y(h,0)=0$ for all $h$ (the formula gives $0$), so: $f_{yx}(0,0)=\lim_{h\to0}\frac{0-0}{h}=\mathbf{0}.$

Continuity: $f_{xy}(x,y)=y^4(5x+y^2)/(x+y^2)^3$ off origin. Along $y=0$: limit is $0\ne1$. $f_{xy}$ is discontinuous. Both $f_{xy}$ and $f_{yx}$ are discontinuous at $(0,0)$.

2025 Paper 1, 2025-P1-Q4b (15 marks)

$f(x,y)=xy(x^2-y^2)/(x^2+y^2)$ for $(x,y)\ne(0,0)$, $f(0,0)=0$. Find $f_{xy}(0,0)$ and $f_{yx}(0,0)$.

$f_x(0,y)$: $[f(h,y)-0]/h=y(h^2-y^2)/(h^2+y^2)\to y(-y^2/y^2)=-y$. So $f_x(0,y)=-y$: $f_{xy}(0,0)=\lim_{k\to0}\frac{-k}{k}=\mathbf{-1}.$

$f_y(x,0)$: $[f(x,k)-0]/k=x(x^2-k^2)/(x^2+k^2)\to x$. So $f_y(x,0)=x$: $f_{yx}(0,0)=\lim_{h\to0}\frac{h}{h}=\mathbf{+1}.$

$\boxed{f_{xy}(0,0)=-1,\quad f_{yx}(0,0)=+1.}$

Common Traps

• Never plug $(0,0)$ into the formula for $f_x$ or $f_y$ — the formula is $0/0$ at the origin. The limit definition is mandatory.
• The formula for $f_{xy}$ off the origin may look the same as $f_{yx}$, but their values at the origin are determined by separate iterated limits, not by the formula.
• For the continuity discussion, one contradiction suffices: approach along $y=0$ typically gives one limit, the value $f_{xy}(0,0)$ is another.

euler-homogeneous (2 question(s); 2019, 2025)

Recognition Cues

• A function $u=\phi^{-1}(g(x,y))$ where $g$ involves fractional powers; “show $g$ is homogeneous of degree $n$, hence show [second-order PDE].”
• “If $u=xf(y/x)+g(y/x)$, show [first-order Euler identity] and [second-order Euler identity].”

Solution Template

1. Identify the homogeneous piece. Not $u$ itself, but $v=\psi(u)$ (e.g. $\sin^2u$). Check: does $v(tx,ty)=t^n v(x,y)$? If yes, $n$ is the degree.
2. First-order Euler. $xv_x+yv_y=nv$. Use chain rule $v_x=\psi'(u)u_x$: $xu_x+yu_y=nv/\psi'(u)=:G(u)$.
3. Second-order Euler. Differentiate $xu_x+yu_y=G(u)$ w.r.t. $x$ (multiply by $x$) and w.r.t. $y$ (multiply by $y$); add: $x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=G(u)[G'(u)-1]$.
4. Substitute $G(u)$ and $G'(u)$ for the specific $\psi$.

Worked Example(s)

2019 Paper 1, 2019-P1-Q4c-i (12 marks)

$u=\arcsin\sqrt{(x^{1/3}+y^{1/3})/(x^{1/2}+y^{1/2})}$. Show $\sin^2u$ has degree $-1/6$. Hence derive the second-order identity.

$v=\sin^2u=(x^{1/3}+y^{1/3})/(x^{1/2}+y^{1/2})$. $v(tx,ty)=t^{1/3-1/2}v=t^{-1/6}v$, so $n=-1/6$. ✓

$G(u)=nv/\psi'(u)=(-\tfrac16\sin^2u)/(2\sin u\cos u)=-\tfrac{1}{12}\tan u$.

$G'(u)=-\tfrac{1}{12}\sec^2u$. $G'(u)-1=-\tfrac{13}{12}-\tfrac{\tan^2u}{12}$.

$\boxed{x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=\frac{\tan u}{12}\!\left(\frac{13}{12}+\frac{\tan^2u}{12}\right).}$

2025 Paper 1, 2025-P1-Q3c-ii (10 marks)

$u=xf(y/x)+g(y/x)$. Show (I) $xu_x+yu_y=xf(y/x)$; (II) $x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=0$.

Write $u=P+Q$: $P=xf(y/x)$ homogeneous degree 1, $Q=g(y/x)$ homogeneous degree 0.

Euler: $xP_x+yP_y=1\cdot P=xf(y/x)$; $xQ_x+yQ_y=0$. Sum: $xu_x+yu_y=xf(y/x)$. ✓

Second order ($n(n-1)=0$ for both $n=1$ and $n=0$): $x^2P_{xx}+\ldots=0$ and $x^2Q_{xx}+\ldots=0$. Sum: $x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=0$. ✓

Common Traps

• It is rarely $u$ itself that is homogeneous — look for a composite $\psi(u)$.
• The second-order formula contains a $-1$ in $[G'(u)-1]$ from the $xu_x+yu_y$ term that appears when you differentiate the first-order identity. Dropping this $-1$ produces a wrong coefficient in the final PDE.
• For the $u=xf+g$ type, the elegant route is Euler on each homogeneous part separately; expanding everything with the chain rule and $t=y/x$ also works but takes longer.

partial-existence (1 question(s); 2018)

Recognition Cues

• Piecewise $f$ with seam along $y=0$ or $x=0$; “which of $f_x(a,b)$ and $f_y(a,b)$ exists?”
• The specified point $(a,b)$ may lie in the interior of one branch (making both partials exist trivially) or on the seam.

Solution Template

1. Check the branch. Is $(a,b)$ strictly inside $\{y>0\}$, $\{y<0\}$, or on $y=0$?
2. If strictly interior: both partials exist and equal the derivatives of that branch’s formula at $(a,b)$.
3. If on the seam: compute the one-sided limits of the difference quotient in the perpendicular direction.

Worked Example(s)

2018 Paper 1, 2018-P1-Q3b (12 marks)

$f(x,y)=xy^2$ if $y>0$; $-xy^2$ if $y\le0$. Which of $f_x(0,1)$ and $f_y(0,1)$ exists?

$(0,1)$ has $y=1>0$, so the point is interior to the $y>0$ branch.

$f_x(0,1)=\lim_{h\to0}(h\cdot1-0)/h=1$. Exists. $\quad f_y(0,1)=\lim_{k\to0}(0\cdot(1+k)^2-0)/k=0$. Exists.

Both exist. The seam at $y=0$ is irrelevant for a point at $y=1$.

Common Traps

• The seam is a distractor unless the point is actually on it. Always check the coordinates first.
• If the point is $(a,0)$ on the seam and $a\ne0$, then $f_y(a,0)$ requires a one-sided limit argument; $f_x(a,0)$ typically exists by comparing left and right limits in $x$.

differential-approximation (1 question(s); 2019)

Recognition Cues

• “Using differentials, find an approximate value of $f(a+\epsilon,b+\delta)$.”
• The base point $(a,b)$ is chosen so $f(a,b)$ is a round number.

Solution Template

1. Choose base $(x_0,y_0)$ where $f$ is clean; compute $\Delta x,\Delta y$.
2. Compute $f_x(x_0,y_0)$ and $f_y(x_0,y_0)$ exactly.
3. Apply $f\approx f_0+f_x\Delta x+f_y\Delta y$.

Worked Example(s)

2019 Paper 2, 2019-P2-Q2c (15 marks)

Approximate $f(4.1,4.9)$ where $f(x,y)=(x^3+x^2y)^{1/2}$.

Base $(4,5)$: $f(4,5)=\sqrt{144}=12$. $\Delta x=0.1$, $\Delta y=-0.1$.

$f_x(4,5)=(48+40)/24=11/3$; $f_y(4,5)=16/24=2/3$.

$df=\tfrac{11}{3}(0.1)+\tfrac23(-0.1)=0.3$. So $\boxed{f(4.1,4.9)\approx12.3.}$

Common Traps

• $\Delta y=4.9-5=-0.1$: the two terms partially cancel. A sign error gives $df=0.433$ instead of $0.3$.
• Keep exact fractions for the partial derivatives; rounding $11/3\approx3.67$ and $2/3\approx0.67$ then multiplying introduces avoidable error.

Practice Set

• 2017-P1-Q3c (15 m) — — mixed partials at origin for a different piecewise function; apply the limit-definition template.
• 2021-P1-Q2b (15 m) — — $f=|x^2-y^2|$; compute $f_{xy}(0,0)$ and $f_{yx}(0,0)$ (they turn out to be equal here).