Exercises — Chain rule for multivariable functions — all cases

There are two paths to $t$: one through $x$, one through $y$. Multiply along each, add: $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$ We write $d$ on the left because after substituting $x(t),y(t)$, the output $z$ depends on the single ultimate variable $t$. The two inside pieces stay $\partial$ because $z$ genuinely has two inputs $x,y$.

Three inner functions $\Rightarrow$ three paths from $w$ down to $t$: $\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}.$ All symbols are $\partial$: the output survives depending on two ultimate variables $s$ and $t$.

Now $t$ has three roads: through $x$, through $y$, and the direct explicit slot in $f$: $\frac{dw}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial t}.$ The final term $\partial f/\partial t$ means "hold $x,y$ fixed, wiggle only the explicit $t$." Missing it is the classic slip.

Link derivatives. $\frac{\partial z}{\partial x}=2xy$, $\frac{\partial z}{\partial y}=x^2+3$. Then $\frac{dx}{dt}=2t$, $\frac{dy}{dt}=e^t$. Assemble (two paths, add): $\frac{dz}{dt}=2xy\cdot 2t+(x^2+3)e^t.$ Substitute $x=t^2,y=e^t$: $\frac{dz}{dt}=2t^2{e}^t\cdot 2t+(t^4+3)e^t=4t^3{e}^t+(t^4+3)e^t.$ At $t=0$: $x=0,y=1$, so $\frac{dz}{dt}{|}_0=0+(0+3)\cdot 1=\boxed{3}.$

$\frac{\partial z}{\partial x}=\frac{2x}{x^2+y^2}$, $\frac{\partial z}{\partial y}=\frac{2y}{x^2+y^2}$. And $\frac{\partial x}{\partial \theta }=-r\sin \theta$, $\frac{\partial y}{\partial \theta }=r\cos \theta$. $\frac{\partial z}{\partial \theta }=\frac{2x}{x^2+y^2}(-r\sin \theta )+\frac{2y}{x^2+y^2}(r\cos \theta ).$ Since $x^2+y^2=r^2$, substitute $x=r\cos \theta ,y=r\sin \theta$: $=\frac{2r\cos \theta (-r\sin \theta )+2r\sin \theta (r\cos \theta )}{r^2}=\frac{-2r^2\cos \theta \sin \theta +2r^2\sin \theta \cos \theta }{r^2}=\boxed{0}.$ Geometry: $z=\ln (r^2)$ depends only on distance $r$, not on the angle — rotating around the origin doesn't change $z$, so its $\theta$-rate is zero. The rule "discovers" this fact.

$\frac{\partial w}{\partial x}=y,\frac{\partial w}{\partial y}=x,\frac{\partial w}{\partial z}=1$. Inner rates: $1,2t,3t^2$. $\frac{dw}{dt}=y\cdot 1+x\cdot 2t+1\cdot 3t^2=t^2+t\cdot 2t+3t^2=t^2+2t^2+3t^2=6t^2.$ At $t=2$: $6(4)=\boxed{24}$. (Check: $w=t\cdot t^2+t^3=2t^3$, so $dw/dt=6t^2$. ✓)

This is the "implicit reappearance" case — $t$ has three paths. $\frac{\partial f}{\partial x}=2x,\frac{\partial f}{\partial y}=2y,\frac{\partial f}{\partial t}=1$. Inner: $\frac{dx}{dt}=-\sin t,\frac{dy}{dt}=\cos t$. $\frac{dw}{dt}=2x(-\sin t)+2y(\cos t)+1.$ Substitute $x=\cos t,y=\sin t$: $=-2\cos t\sin t+2\sin t\cos t+1=0+1=\boxed{1}.$ Interpretation: the point $(x,y)$ runs around the unit circle, so $x^2+y^2=1$ never changes — those two paths cancel. Only the explicit $+t$ term survives, giving rate $1$. Drop $\partial f/\partial t$ and you'd get the wrong answer $0$.

$\frac{\partial u}{\partial x}=1,\frac{\partial v}{\partial x}=-1$, and $\frac{\partial u}{\partial y}=-1,\frac{\partial v}{\partial y}=1$. $\frac{\partial z}{\partial x}=f_u(1)+f_v(-1)=f_u-f_v,$ $\frac{\partial z}{\partial y}=f_u(-1)+f_v(1)=-f_u+f_v.$ Add: $(f_u-f_v)+(-f_u+f_v)=\boxed{0}$. Why: $v=-u$, so $z$ really depends on one combination $x-y$; moving $x$ and $y$ by equal amounts keeps $x-y$ fixed, hence no change.

Inner rates: $x_s=e^s\cos t=x$, $y_s=e^s\sin t=y$, $x_t=-e^s\sin t=-y$, $y_t=e^s\cos t=x$. $z_s=z_{x}x+z_{y}y,z_t=z_x(-y)+z_{y}x=-z_{x}y+z_{y}x.$ Square and add: $z_s^2+z_t^2=(z_{x}x+z_{y}y{)}^2+(-z_{x}y+z_{y}x{)}^2.$ Expand: $z_x^2{x}^2+2z_x{z}_{y}xy+z_y^2{y}^2+z_x^2{y}^2-2z_x{z}_{y}xy+z_y^2{x}^2=z_x^2(x^2+y^2)+z_y^2(x^2+y^2)$. Since $x^2+y^2=e^{2s}$: $\boxed{z_s^2+z_t^2=e^{2s}\left(z_x^2+z_y^2\right)}.$

Pure bookkeeping — no formula for $f$ needed: $\frac{dz}{dt}=f_x{x}_t+f_y{y}_t=(3)(2)+(-1)(5)=6-5=\boxed{1}.$

Inner rates: $x_s=1,y_s=1,x_t=1,y_t=-1$. $z_s=z_x(1)+z_y(1)=z_x+z_y,z_t=z_x(1)+z_y(-1)=z_x-z_y.$ Multiply — difference of squares: $z_s{z}_t=(z_x+z_y)(z_x-z_y)=z_x^2-z_y^2.\boxed{✓}$

First derivative (from the parent note): $z_r=f_x\cos \theta +f_y\sin \theta$. Differentiate again w.r.t. $r$, holding $\theta$ fixed. The trig factors $\cos \theta ,\sin \theta$ are constants in $r$; but $f_x$ and $f_y$ are still functions of $x,y$, hence of $r$. Apply the chain rule to each: $\frac{\partial f_x}{\partial r}=f_{xx}\cos \theta +f_{xy}\sin \theta ,\frac{\partial f_y}{\partial r}=f_{yx}\cos \theta +f_{yy}\sin \theta .$ So $z_{rr}=\cos \theta (f_{xx}\cos \theta +f_{xy}\sin \theta )+\sin \theta (f_{xy}\cos \theta +f_{yy}\sin \theta ),$ using $f_{yx}=f_{xy}$ (equality of mixed partials). Collect: $\boxed{z_{rr}=f_{xx}{\cos }^2\theta +2f_{xy}\cos \theta \sin \theta +f_{yy}{\sin }^2\theta }.$

Let $u=\lambda x$, $v=\lambda y$, and define $g(\lambda )=f(u,v)=f(\lambda x,\lambda y)$. Differentiate both sides in $\lambda$. Left side by the chain rule (two paths, through $u$ and $v$): $g^{\prime }(\lambda )=f_u\frac{du}{d\lambda }+f_v\frac{dv}{d\lambda }=f_{u}x+f_{v}y.$ Right side: $\frac{d}{d\lambda }({\lambda }^{k}f(x,y))=k{\lambda }^{k-1}f(x,y).$ These are equal for every $\lambda$. Set $\lambda =1$, so $u=x,v=y$: $x{f}_x+y{f}_y=kf.\boxed{✓}$ The trick: introduce the scalar knob $\lambda$, differentiate through it, then freeze it at $1$.

Apply the scalar chain rule to each of the four outputs: $u_s=f_x{x}_s+f_y{y}_s,u_t=f_x{x}_t+f_y{y}_t,$ $w_s=g_x{x}_s+g_y{y}_s,w_t=g_x{x}_t+g_y{y}_t.$ Now read the right-hand product entry-by-entry. Row 1 of $J_f$ dotted with column 1 of $J_x$ gives $f_x{x}_s+f_y{y}_s=u_s$; column 2 gives $u_t$; row 2 gives $w_s,w_t$. Every entry matches, so $J_{(u,w)}=J_f{J}_x.\boxed{✓}$ This is why the parent note calls the general chain rule just matrix multiplication of Jacobians — see Gradient and Jacobian.

At $(s,t)=(1,2)$: $x=3,y=2$. Jacobians. $J_f=\left(\begin{array}{cc}2x& 1\\ y& x\end{array}\right)=\left(\begin{array}{cc}6& 1\\ 2& 3\end{array}\right)$. $J_x=\left(\begin{array}{cc}{x}_s& x_t\\ y_s& y_t\end{array}\right)=\left(\begin{array}{cc}1& 1\\ t& s\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right).$ Product top-left entry ($u_s$): $6\cdot 1+1\cdot 2=8$. Direct: $u=x^2+y=(s+t{)}^2+st$, so $u_s=2(s+t)+t=2(3)+2=\boxed{8}$. ✓