06. 편도함수와 그 계산Partial Derivatives and Their Computation

Solution. Since $f_x=\dfrac{\partial f}{\partial x}=2x+3y$, we have $f_x(4,-5)=2(4)+3(-5)=8-15=-7$.

Since $f_y=\dfrac{\partial f}{\partial y}=3x+1$, we have $f_y(4,-5)=3(4)+1=13$.

Solution.

(1) Since $f(x,y)=y\sin xy$, the product rule gives

(2) We have $f(x,y)=\dfrac{2y}{y+\cos x}$. Differentiating with respect to $x$ (quotient rule; the numerator does not contain $x$, so its derivative is $0$),

and differentiating with respect to $y$,

(3) Since $f(x,y,z)=x\sin(y+3z)$,

Solution. Differentiating both sides of $yz-\ln z=x+y$ with respect to $x$ (treating $y$ as constant, $z=z(x,y)$),

Rearranging the left side,

so

Solution. Along the curve of intersection, $x=1$ is fixed, so $z=1+y^2$. Hence the instantaneous rate of change of $z$ with respect to $y$ along this curve is

and since this is the cross-section with $x$ held at the constant $1$, it agrees with $\partial z/\partial y$. At the point $(1,2,5)$ we have $y=2$, so the rate of change is

Solution. With $R_1=30,R_2=45,R_3=90$, we first find the value of $R$:

so $R=15\,\Omega$.

Differentiating both sides of $\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$ with respect to $R_2$ ($R_1,R_3$ constant, $R=R(R_1,R_2,R_3)$),

so

Proof.

(1) Consider points $(x,0)$ on the $x$-axis near the origin, $x\neq0$. Since $xy=0$ there, $f(x,0)=1$ for all $x$. Hence along any sequence $(x_n,0)\to(0,0)$ with $x_n\neq0$, $f(x_n,0)=1\to1$.

On the other hand, consider points $(x,x)$ on the line $y=x$ with $x\neq0$. Then $xy=x^2\neq0$, so $f(x,x)=0$. Hence along any sequence $(x_n,x_n)\to(0,0)$ with $x_n\neq0$, $f(x_n,x_n)=0\to0$.

Since the limits along these two paths are $1$ and $0$, respectively, and they differ, $\displaystyle\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist. Therefore $f$ is not continuous at $(0,0)$.

(2) We compute the partial derivatives directly from the definition. Fixing $y=0$, we have $xy=0$, so $f(h,0)=1$ for every $h\neq0$, and $f(0,0)=1$. Hence

exists and equals $0$. Similarly, fixing $x=0$, we have $f(0,h)=1$ for every $h\neq0$ and $f(0,0)=1$, so

exists and equals $0$. That is, $\partial f/\partial x(0,0)=\partial f/\partial y(0,0)=0$, so both partial derivatives exist at the origin. (This shows that existence of the partial derivatives does not imply continuity.)

Solution. First we find the first-order partial derivatives:

(1) $\dfrac{\partial^2 f}{\partial x^2}=\dfrac{\partial}{\partial x}(f_x)=ye^x$.

(2) $\dfrac{\partial^2 f}{\partial y\,\partial x}=\dfrac{\partial}{\partial y}(f_x)=-\sin y+e^x$.

(3) $\dfrac{\partial^2 f}{\partial y^2}=\dfrac{\partial}{\partial y}(f_y)=-x\cos y$.

(4) $\dfrac{\partial^2 f}{\partial x\,\partial y}=\dfrac{\partial}{\partial x}(f_y)=-\sin y+e^x$. (Since the mixed partials are continuous, the values in (2) and (4) agree.)

Solution. Differentiating with respect to $x$ first (the term $e^y/(y^2+1)$ contains no $x$, so it drops out),

Differentiating this again with respect to $y$,

Solution. $f_{yxyz}$ means differentiating in the order given by the subscripts, i.e. $y\to x\to y\to z$.

Differentiating with respect to $y$ first,

then with respect to $x$,

then again with respect to $y$,

and finally with respect to $z$,

Solution. For $(x,y)\neq(0,0)$, the quotient rule gives

At the origin these formulas are of the form $0/0$, so the values must be found from the definition. When $y=0$, the numerator $xy(x^2-y^2)=0$ (and the denominator $x^2\neq0$ for $x\neq0$), so $f(x,0)=0$ for all $x$; likewise $f(0,y)=0$ for all $y$. Hence

Computing $\dfrac{\partial^2 f}{\partial x\,\partial y}(0,0)$. This is $\dfrac{\partial}{\partial x}(f_y)$ evaluated at the origin. Setting $y=0$, for $x\neq0$ we get $f_y(x,0)=\dfrac{x^5}{x^4}=x$, and $f_y(0,0)=0$, so $f_y(x,0)=x$ for all $x$. Hence

Computing $\dfrac{\partial^2 f}{\partial y\,\partial x}(0,0)$. This is $\dfrac{\partial}{\partial y}(f_x)$ evaluated at the origin. Setting $x=0$, for $y\neq0$ we get $f_x(0,y)=\dfrac{-y^5}{y^4}=-y$, and $f_x(0,0)=0$, so $f_x(0,y)=-y$ for all $y$. Hence

Thus $\dfrac{\partial^2 f}{\partial x\,\partial y}(0,0)=1\neq-1=\dfrac{\partial^2 f}{\partial y\,\partial x}(0,0)$: the two mixed partial derivatives disagree at the origin. (This is the standard example showing that when the mixed partials are not continuous at a point, the conclusion of Clairaut's theorem can fail.)

Solution. Let $r=\sqrt{x^2+y^2+z^2}$, so $f=r^{-1}$. Since $\dfrac{\partial r}{\partial x}=\dfrac{x}{r}$, the chain rule gives

Differentiating again with respect to $x$,

By symmetry in $r$, the same computation for $y$ and $z$ gives

Adding the three expressions,

Solution.

(1) By the Chain Rule,

At $t=\pi/2$, $\cos2t=\cos\pi=-1$, so $\left.\dfrac{dw}{dt}\right|_{t=\pi/2}=-1$.

(2) By the Chain Rule,

Substituting $z=2r$,

Also

Solution.

(1) By the Chain Rule,

(equivalently $=2\cos^2t$).

(2) By the Chain Rule,

Solution. Let $u=s^3+t^2$, so $w=f(u)$ and $f'(u)=e^u$. By the Chain Rule,

Solution. Since $z=f(x,y)^2$ with $x=g(t)$, $y=h(t)$, the Chain Rule gives

At $t=0$ we have $x=g(0)=2$, $y=h(0)=-1$. Substituting the given values,

Solution. Differentiating both sides of $V=IR$ with respect to $t$ (product rule),

Substituting $R=600\,\Omega$, $I=0.04\,\text{A}$, $dR/dt=0.5\,\Omega/\text{s}$, $dV/dt=-0.01\,\text{V/s}$,

so

Solution.

(1) Let $F(x,y)=y^2-x^2-\sin xy$, so $F(x,y)=0$ defines $y$ implicitly as a function of $x$. Computing $F_x,F_y$,

so by implicit differentiation,

(2) Let $F(x,y,z)=x^3+z^2+ye^{xz}+z\cos y$. Then

At the origin $(0,0,0)$, $F(0,0,0)=0+0+0+0=0$, so the origin lies on the surface, and there

Hence

Solution. Let $F(x,y,z)=\sin(x+y)+\sin(y+z)+\sin(z+x)$. Then

Checking that $(\pi,\pi,\pi)$ lies on the surface: $\sin2\pi+\sin2\pi+\sin2\pi=0$, so it does. At this point $x+y=y+z=z+x=2\pi$, so $\cos(x+y)=\cos(y+z)=\cos(z+x)=1$, and

Hence

Solution.

(1) If $x,y$ are independent variables, then $z=z(x,y)=x^2+y^2$, so by the Chain Rule,

Substituting $z=x^2+y^2$,

(2) If $x,z$ are independent variables, then from $z=x^2+y^2$ we get $y^2=z-x^2$. Substituting into $w=x^2+y^2+z^2$,

so $w$ contains no $x$ at all and is a function of $z$ alone. Hence, regarding $x,z$ as independent,

Solution. Let $G(x,y,z)=z^3-xy+yz+y^3-1=0$; since $x,y$ are independent variables, $z=z(x,y)$. Differentiating $G=0$ with respect to $x$ ($y$ fixed),

so

At $(x,y,z)=(2,-1,1)$,

Since $w=x^2+y^2+z^2$ with $x,y$ independent,

Substituting $(x,y,z)=(2,-1,1)$ and $\partial z/\partial x=-1/2$,

Solution. Take $x,y,z$ as independent variables with $t=x+y$. Differentiating $w=x^2+y-z+\sin t$ with respect to $x$, holding $y,z$ fixed (applying the Chain Rule since $t$ depends on $x$),

Since $t=x+y$ with $y$ fixed gives $\left(\partial t/\partial x\right)_y=1$,

Solution. The relation $PV=nRT$ means that once two of $P,V,T$ are taken as independent variables, the remaining one is determined by them. Although $U=f(P,V,T)$ is defined as a function of all three variables, the meaning of a partial derivative depends on which two variables are chosen as independent.

Computing $(\partial U/\partial P)_V$. Holding $V$ fixed and taking $P$ as the independent variable, the relation gives $T=\dfrac{PV}{nR}$, so $T$ depends on $P$ (with $V$ fixed). Applying the Chain Rule to $U=f(P,V,T(P,V))$,

Since $T=PV/(nR)$ with $V$ fixed gives $\left(\partial T/\partial P\right)_V=V/(nR)$,

Computing $(\partial U/\partial T)_V$. Now holding $V$ fixed and taking $T$ as the independent variable, the relation gives $P=\dfrac{nRT}{V}$, so $P$ depends on $T$ (with $V$ fixed). Applying the Chain Rule in the same way,

Since $P=nRT/V$ with $V$ fixed gives $\left(\partial P/\partial T\right)_V=nR/V$,