Exercises — Partial derivatives — notation, calculation, geometric meaning

What we do: differentiate $x$-first, treating $y$ as a number.

• $f_x$: $5x^2\to 10x$; the term $3y$ has ==no $x$== so it is a constant $\Rightarrow 0$; $-8\to 0$. $f_x=10x$
• $f_y$: now $5x^2$ is a frozen constant $\Rightarrow 0$; $3y\to 3$; $-8\to 0$. $f_y=3$

• $f_x$: $y$ is a constant multiplier. $\frac{d}{dx}{x}^3=3x^2$, keep the $y$: $f_x=3x^{2}y$
• $f_y$: $x^3$ is the constant multiplier. $\frac{d}{dy}y=1$: $f_y=x^3$

What it looks like: a perfectly flat surface — no slope in any direction. $f_x=0,f_y=0,f_z=0$ Why: the derivative of a constant is $0$ no matter which variable you wiggle.

• $f_x$: $e^y$ is a constant coefficient. $\frac{d}{dx}\sin x=\cos x$: $f_x=\cos (x)e^y$
• $f_y$: $\sin x$ is the constant coefficient. $\frac{d}{dy}{e}^y=e^y$: $f_y=\sin (x)e^y$

Which tool and why? The exponent $x^{2}y$ is itself a function, so we need the chain rule: derivative of $e^{(\text{inner})}$ is $e^{(\text{inner})}\cdot (\text{inner}{)}^{\prime }$.

• $f_x$: inner $=x^{2}y$, its $x$-derivative is $2xy$ ($y$ frozen): $f_x=e^{x^{2}y}\cdot 2xy$
• $f_y$: inner's $y$-derivative is $x^2$ ($x$ frozen): $f_y=e^{x^{2}y}\cdot x^2$

Which tool and why? A ratio of two things both containing the variable → quotient rule ${\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$.

• $f_x$: $u=x,u_x=1$; $v=x+y,v_x=1$: $f_x=\frac{1\cdot (x+y)-x\cdot 1}{(x+y{)}^2}=\frac{y}{(x+y{)}^2}$
• $f_y$: now $u=x$ is a constant so $u_y=0$; $v=x+y,v_y=1$: $f_y=\frac{0\cdot (x+y)-x\cdot 1}{(x+y{)}^2}=\frac{-x}{(x+y{)}^2}$

Differentiate first, evaluate last (never the reverse). $f_x=2x\Rightarrow f_x(3,-1)=6,f_y=2y\Rightarrow f_y(3,-1)=-2$ Steepness is measured by $|\text{slope}|$: $|6|>|-2|$, so the surface is steeper in the $x$-direction here. The negative $f_y$ means moving in $+y$ makes the height decrease (we're at $y=-1$, on the down-slope side of the bowl).

Chain rule: $\frac{d}{dx}\ln (u)=\frac{u_x}{u}$ with $u=x^2+y^2$. $f_x=\frac{2x}{x^2+y^2},f_y=\frac{2y}{x^2+y^2}$ At $(1,0)$: $f_x=\frac{2}{1}=2$, $f_y=\frac{0}{1}=0$. Degenerate case $(0,0)$: $x^2+y^2=0$ there, so both partials have $\frac{0}{0}$ — undefined. Also $\ln 0$ is $-\infty$, so $f$ itself isn't even defined at the origin. The partials exist everywhere except that single hole.

Formula (from parent): $z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$.

• $f(1,2)=1^2\cdot 2^3=8$.
• $f_x=2x{y}^3\Rightarrow f_x(1,2)=2\cdot 1\cdot 8=16$.
• $f_y=3x^2{y}^2\Rightarrow f_y(1,2)=3\cdot 1\cdot 4=12$. $\boxed{z=8+16(x-1)+12(y-2)}$ The two slice-slopes $16$ and $12$ are exactly the plane's tilt in $x$ and $y$ — see the figure below.

Step 1 — $f_x$: ($y$ frozen) $f_x=3x^2{y}^2+\cos (xy)\cdot y$. Step 2 — $f_{xy}={\partial }_y(f_x)$: differentiate that in $y$.

• $3x^2{y}^2\to 6x^{2}y$.
• $\cos (xy)y$: product of $\cos (xy)$ and $y$, both depend on $y$. Product rule: ${\partial }_y[\cos (xy)]\cdot y+\cos (xy)\cdot 1=-\sin (xy)\cdot x\cdot y+\cos (xy)$ $f_{xy}=6x^{2}y-xy\sin (xy)+\cos (xy)$ Step 3 — the other order. $f_y=2x^{3}y+\cos (xy)\cdot x$, then $f_{yx}={\partial }_x(f_y)$:
• $2x^{3}y\to 6x^{2}y$.
• $\cos (xy)x$: product rule in $x$: $-\sin (xy)\cdot y\cdot x+\cos (xy)$. $f_{yx}=6x^{2}y-xy\sin (xy)+\cos (xy)$ They match ✔ — as Clairaut's theorem guarantees, since these second partials are continuous.

From Problem 3.2, $f_x=\frac{2x}{x^2+y^2}$. Differentiate again in $x$ (quotient rule, $u=2x$, $v=x^2+y^2$): $f_{xx}=\frac{2(x^2+y^2)-2x\cdot 2x}{(x^2+y^2{)}^2}=\frac{2y^2-2x^2}{(x^2+y^2{)}^2}$ By the symmetry $x\leftrightarrow y$ (swap roles), $f_{yy}=\frac{2x^2-2y^2}{(x^2+y^2{)}^2}$. Add: $f_{xx}+f_{yy}=\frac{(2y^2-2x^2)+(2x^2-2y^2)}{(x^2+y^2{)}^2}=\frac{0}{(x^2+y^2{)}^2}=0✓$

Why two different tools? The variable sits in the base for $f_x$ (power rule) but in the exponent for $f_y$ (exponential rule) — the same expression needs different rules depending on which slot you wiggle.

• $f_x$: treat $y$ as a fixed exponent → power rule $\frac{d}{dx}{x}^y=y{x}^{y-1}$. $f_x=y{x}^{y-1}$
• $f_y$: treat $x$ as a fixed base. Rewrite $x^y=e^{y\ln x}$; then ${\partial }_y=e^{y\ln x}\cdot \ln x=x^y\ln x$. $f_y=x^y\ln x$

Which tool and why? We're asked to bypass the rules and use the definition — this is what the rules are shorthand for. $f_x(2,3)={\lim }_{h\to 0}\frac{f(2+h,3)-f(2,3)}{h}$ Compute the pieces ($y=3$ frozen):

• $f(2+h,3)=(2+h{)}^2\cdot 3=3(4+4h+h^2)=12+12h+3h^2$.
• $f(2,3)=4\cdot 3=12$. Numerator $=12h+3h^2$. Divide by $h$: $\frac{12h+3h^2}{h}=12+3h\underset{h\to 0}{\to }12$ $f_x(2,3)=12$ Check with the rule: $f_x=2xy\Rightarrow 2\cdot 2\cdot 3=12$ ✔.

Why the definition, not the rules? At the origin the formula changes; the quotient rule doesn't apply to the stitched point. Only the limit sees both pieces. $f_x(0,0)={\lim }_{h\to 0}\frac{f(h,0)-f(0,0)}{h}$ Now $f(h,0)=\frac{h\cdot 0}{h^2+0}=0$ and $f(0,0)=0$, so the quotient is $\frac{0-0}{h}=0$ for every $h\ne 0$: $f_x(0,0)=0$ By the identical computation along $y$ (set $x=0$): $f(0,k)=\frac{0\cdot k}{0+k^2}=0$, so $f_y(0,0)=0$ The punchline: both partials exist and equal $0$, yet this $f$ is famously not continuous at the origin (approach along $y=x$ gives $\frac{1}{2}$, along the axes gives $0$). So partials existing does not guarantee smoothness — a deep warning for Tangent plane and linear approximation.

Write $f=(x^2+y^2+z^2{)}^{-1/2}$. Chain rule with inner $u=x^2+y^2+z^2$, $u_x=2x$: $f_x=-\frac{1}{2}(x^2+y^2+z^2{)}^{-3/2}\cdot 2x=\frac{-x}{(x^2+y^2+z^2{)}^{3/2}}$ At $(1,2,2)$: denominator base $=1+4+4=9$, and $9^{3/2}=(3^2{)}^{3/2}=3^3=27$. So $f_x(1,2,2)=\frac{-1}{27}$

Variable in the base ($x^y$, find $f_x$) ::: power rule → $y{x}^{y-1}$. Variable in the exponent ($x^y$, find $f_y$) ::: rewrite $e^{y\ln x}$ → $x^y\ln x$. Both factors depend on the variable ::: product rule (two terms). Ratio, both parts depend on variable ::: quotient rule. Point at origin of a piecewise / singular $f$ ::: go back to the limit definition.