§14.5 연쇄법칙The Chain Rule

Solution. Since $x$ and $y$ are both functions of the single variable $t$, Case 1 applies. Here $z_x=2xy+3y^4$, $z_y=x^2+12xy^3$, while $dx/dt=2\cos 2t$ and $dy/dt=-\sin t$, so

At $t=0$, $x=\sin 0=0$ and $y=\cos 0=1$. Substituting, the first bracket becomes $2(0)(1)+3(1)^4=3$ and the second becomes $0^2+12(0)(1)^3=0$, so

Hence $dz/dt|_{t=0}=6$.

Solution. From $PV=8.31T$, $P=8.31T/V$, and since both $T$ and $V$ are functions of time $t$, Case 1 gives

At the instant in question $T=300$, $dT/dt=0.1$, $V=100$, $dV/dt=0.2$, so

The value is negative, so at that instant the pressure is falling at $0.04155$ kPa/s, about $0.042$ kPa/s.

Solution. Since $x$ and $y$ are both functions of $s$ and $t$, Case 2 applies. Here $z_x=e^x\sin y$, $z_y=e^x\cos y$, and from $x=st^2$, $x_s=t^2$, $x_t=2st$, while from $y=s^2t$, $y_s=2st$, $y_t=s^2$, so

Substituting $x=st^2$ and $y=s^2t$ to express the result in terms of $s$ and $t$ alone gives

Solution. This is the general Chain Rule with $n=4$ intermediate variables ($x,y,z,t$) and $m=2$ independent variables ($u,v$). The tree above, with branches from the dependent variable $w$ to each intermediate variable and from each of those four to $u$ and to $v$, shows there are four paths from $w$ to $u$ and four from $w$ to $v$. Theorem 3 sums the product of partial derivatives along each path:

Each equation has four terms, one per intermediate variable.

Solution. There are three intermediate variables $x,y,z$ and three independent variables $r,s,t$. As the highlighted branches in the tree diagram above show, the paths from $u$ to $s$ are $u\to x\to s$, $u\to y\to s$ and $u\to z\to s$, and the general Chain Rule adds the product of partial derivatives along each of the three.

The derivatives needed are $\partial u/\partial x=4x^3y$, $\partial u/\partial y=x^4+2yz^3$, $\partial u/\partial z=3y^2z^2$, together with $\partial x/\partial s=re^{t}$, $\partial y/\partial s=2rse^{-t}$, $\partial z/\partial s=r^2\sin t$.

At $r=2,\,s=1,\,t=0$ we have $x=2,\,y=2,\,z=0$, so $\partial u/\partial x=4(2)^3(2)=64$, $\partial u/\partial y=2^4+2(2)(0)^3=16$, $\partial u/\partial z=3(2)^2(0)^2=0$, while $\partial x/\partial s=2e^{0}=2$, $\partial y/\partial s=2(2)(1)e^{0}=4$, $\partial z/\partial s=2^2\sin 0=0$.

Multiplying and adding along the three paths,

Hence $\partial u/\partial s=192$.

Proof. Write $x=s^2-t^2$ and $y=t^2-s^2$, so $g(s,t)=f(x,y)$ with $x,y$ both functions of $s,t$, and Case 2 applies. Since $x_s=2s$, $y_s=-2s$, $x_t=-2t$, $y_t=2t$,

Hence

(a) Since $x$ and $y$ are both functions of $r$ and $s$, Case 2 applies. With $x_r=2r$ and $y_r=2s$,

(b) Differentiate the last expression in $r$ once more. Since $z_x$ and $z_y$ are themselves functions of $x$ and $y$, they too must be differentiated in $r$ via Case 2. Combining the product rule with the Chain Rule,

Treating $\partial z/\partial x$ as a new function and applying Case 2 gives $\dfrac{\partial}{\partial r}\Big(\dfrac{\partial z}{\partial x}\Big)=\dfrac{\partial^2z}{\partial x^2}(2r)+\dfrac{\partial^2z}{\partial y\partial x}(2s)$, and likewise $\dfrac{\partial}{\partial r}\Big(\dfrac{\partial z}{\partial y}\Big)=\dfrac{\partial^2z}{\partial x\partial y}(2r)+\dfrac{\partial^2z}{\partial y^2}(2s)$. Because the second-order partials of $z$ are continuous, Clairaut's Theorem gives $\dfrac{\partial^2z}{\partial y\partial x}=\dfrac{\partial^2z}{\partial x\partial y}$, and substituting this and simplifying leaves

Solution. Let $F(x,y)=x^3+y^3-6xy$, so the given equation reads $F(x,y)=0$. Here $F_x=3x^2-6y$ and $F_y=3y^2-6x$, and formula (5) is available at any point where $F_y\neq0$, that is, where $y^2\neq2x$. At such a point

Hence $y'=-\dfrac{x^2-2y}{y^2-2x}$.

Solution. Let $F(x,y,z)=x^3+y^3+z^3+6xyz+4$, so the given equation reads $F(x,y,z)=0$. Its partial derivatives are $F_x=3x^2+6yz$, $F_y=3y^2+6xz$, $F_z=3z^2+6xy$, and formulas (6) are available wherever $F_z\neq0$, that is, wherever $z^2+2xy\neq0$. At such a point

The same surface was handled in §14.3, Example 5 by differentiating the equation directly; the answers agree, and formulas (6) simply package that computation.