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Partial Derivatives

2.3 Higher Order Derivatives

7 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Let all of the third order partial derivatives of the function f(x,y,z)f(x,y,z) exist and be continuous. Show that

fxyz(x,y,z)=fxzy(x,y,z)=fyxz(x,y,z)=fyzx(x,y,z)=fzxy(x,y,z)=fzyx(x,y,z)\begin{equation*} f_{xyz}(x,y,z) =f_{xzy}(x,y,z) =f_{yxz}(x,y,z) =f_{yzx}(x,y,z) =f_{zxy}(x,y,z) =f_{zyx}(x,y,z) \end{equation*}
Hint

Repeatedly use (Clairaut's) Theorem 2.3.4 in the CLP-3 text.

Answer

See the solution.

Full solution

We have to derive a bunch of equalities.

  • Fix any real number xx and set g(y,z)=fx(x,y,z)g(y,z)=f_x(x,y,z). By (Clairaut's) Theorem 2.3.4 in the CLP-3 text gyz(y,z)=gzy(y,z)g_{yz}(y,z)=g_{zy}(y,z), so

    fxyz(x,y,z)=gyz(y,z)=gzy(y,z)=fxzy(x,y,z)\begin{equation*} f_{xyz}(x,y,z) = g_{yz}(y,z) =g_{zy}(y,z) = f_{xzy}(x,y,z) \end{equation*}

  • For every fixed real number zz, (Clairaut's) Theorem 2.3.4 in the CLP-3 text gives fxy(x,y,z)=fyx(x,y,z)f_{xy}(x,y,z)=f_{yx}(x,y,z). So

    fxyz(x,y,z)=∂∂zfxy(x,y,z)=∂∂zfyx(x,y,z)=fyxz(x,y,z)\begin{equation*} f_{xyz}(x,y,z) = \pdiff{}{z} f_{xy}(x,y,z)= \pdiff{}{z} f_{yx}(x,y,z) =f_{yxz}(x,y,z) \end{equation*}

    So far, we have

    fxyz(x,y,z)=fxzy(x,y,z)=fyxz(x,y,z)\begin{equation*} f_{xyz}(x,y,z) = f_{xzy}(x,y,z)=f_{yxz}(x,y,z) \end{equation*}

  • Fix any real number yy and set g(x,z)=fy(x,y,z)g(x,z)=f_y(x,y,z). By (Clairaut's) Theorem 2.3.4 in the CLP-3 text gxz(x,z)=gzx(x,z)g_{xz}(x,z)=g_{zx}(x,z). So

    fyxz(x,y,z)=gxz(x,z)=gzx(x,z)=fyzx(x,y,z)\begin{equation*} f_{yxz}(x,y,z) = g_{xz}(x,z) =g_{zx}(x,z) = f_{yzx}(x,y,z) \end{equation*}

    So far, we have

    fxyz(x,y,z)=fxzy(x,y,z)=fyxz(x,y,z)=fyzx(x,y,z)\begin{equation*} f_{xyz}(x,y,z) = f_{xzy}(x,y,z)=f_{yxz}(x,y,z)= f_{yzx}(x,y,z) \end{equation*}

  • For every fixed real number yy, (Clairaut's) Theorem 2.3.4 in the CLP-3 text gives fxz(x,y,z)=fzx(x,y,z)f_{xz}(x,y,z)=f_{zx}(x,y,z). So

    fxzy(x,y,z)=∂∂yfxz(x,y,z)=∂∂yfzx(x,y,z)=fzxy(x,y,z)\begin{equation*} f_{xzy}(x,y,z) = \pdiff{}{y} f_{xz}(x,y,z)= \pdiff{}{y} f_{zx}(x,y,z) =f_{zxy}(x,y,z) \end{equation*}

    So far, we have

    fxyz(x,y,z)=fxzy(x,y,z)=fyxz(x,y,z)=fyzx(x,y,z)=fzxy(x,y,z)\begin{equation*} f_{xyz}(x,y,z) = f_{xzy}(x,y,z)=f_{yxz}(x,y,z)= f_{yzx}(x,y,z)=f_{zxy}(x,y,z) \end{equation*}

  • Fix any real number zz and set g(x,y)=fz(x,y,z)g(x,y)=f_z(x,y,z). By (Clairaut's) Theorem 2.3.4 in the CLP-3 text gxy(x,y)=gyx(x,y)g_{xy}(x,y)=g_{yx}(x,y). So

    fzxy(x,y,z)=gxy(x,y)=gyx(x,y)=fzxy(x,y,z)\begin{equation*} f_{zxy}(x,y,z) = g_{xy}(x,y) =g_{yx}(x,y) = f_{zxy}(x,y,z) \end{equation*}

    We now have all of

    fxyz(x,y,z)=fxzy(x,y,z)=fyxz(x,y,z)=fyzx(x,y,z)=fzxy(x,y,z)=fzxy(x,y,z)\begin{equation*} f_{xyz}(x,y,z) = f_{xzy}(x,y,z)=f_{yxz}(x,y,z)= f_{yzx}(x,y,z)=f_{zxy}(x,y,z) = f_{zxy}(x,y,z) \end{equation*}
Q2Stage 1

Find, if possible, a function f(x,y)f(x,y) for which fx(x,y)=eyf_x(x,y)=e^y and fy(x,y)=exf_y(x,y)=e^x.

Hint

If f(x,y)f(x,y) obeying the specified conditions exists, then it is necessary that fxy(x,y)=fyx(x,y)f_{xy}(x,y)=f_{yx}(x,y).

Answer

No such f(x,y)f(x,y) exists.

Full solution

No such f(x,y)f(x,y) exists, because if it were to exist, then we would have that fxy(x,y)=fyx(x,y)f_{xy}(x,y)=f_{yx}(x,y). But

fxy(x,y)=∂∂yfx(x,y)=∂∂yey=eyfyx(x,y)=∂∂xfy(x,y)=∂∂xex=ex\begin{align*} f_{xy}(x,y)&=\pdiff{}{y}f_x(x,y)=\pdiff{}{y}e^y=e^y \\ f_{yx}(x,y)&=\pdiff{}{x}f_y(x,y)=\pdiff{}{x}e^x=e^x \end{align*}

are not equal.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q3Stage 2

Find the specified partial derivatives.

  1. f(x,y)=x2y3f(x,y) = x^2y^3; fxx(x,y)f_{xx}(x,y), fxyy(x,y)f_{xyy}(x,y), fyxy(x,y)f_{yxy}(x,y)

  2. f(x,y)=exy2f(x,y) = e^{xy^2}; fxx(x,y)f_{xx}(x,y), fxy(x,y)f_{xy}(x,y), fxxy(x,y)f_{xxy}(x,y), fxyy(x,y)f_{xyy}(x,y)

  3. f(u,v,w)=1u+2v+3w\displaystyle f(u,v,w) = \frac{1}{u+2v+3w}, ∂3f∂u∂v∂w(u,v,w)\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(u,v,w), ∂3f∂u∂v∂w(3,2,1)\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(3,2,1)

Answer

(a) fxx(x,y)=2y3f_{xx}(x,y) = 2y^3 fyxy(x,y)=fxyy(x,y)=12xyf_{yxy}(x,y) = f_{xyy}(x,y) = 12xy

(b) fxx(x,y)=y4exy2f_{xx}(x,y)= y^4e^{xy^2} fxy(x,y)=(2y+2xy3)exy2f_{xy}(x,y)= \big(2y+2xy^3\big)e^{xy^2} fxxy(x,y)=(4y3+2xy5)exy2f_{xxy}(x,y)= \big(4y^3 + 2xy^5\big)e^{xy^2} fxyy(x,y)=(2+10xy2+4x2y4)exy2f_{xyy}(x,y) = \big(2+10xy^2+4x^2y^4\big)e^{xy^2}

(c) ∂3f∂u ∂v ∂w(u,v,w)=−36(u+2v+3w)4\displaystyle\frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(u,v,w) = -\frac{36}{(u+2v+3w)^4} ∂3f∂u ∂v ∂w(3,2,1)=−0.0036=−92500\displaystyle\frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(3,2,1) = -0.0036 = -\frac{9}{2500}

Full solution

(a) We have

fx(x,y)=2xy3fxx(x,y)=2y3fxy(x,y)=6xy2fyxy(x,y)=fxyy(x,y)=12xy\begin{align*} f_x(x,y) &= 2xy^3 & f_{xx}(x,y) &= 2y^3 \\ & & f_{xy}(x,y) &= 6xy^2 & f_{yxy}(x,y) = f_{xyy}(x,y) &= 12xy \end{align*}

(b) We have

fx(x,y)=y2exy2fxx(x,y)=y4exy2fxxy(x,y)=4y3exy2+2xy5exy2fxy(x,y)=2yexy2+2xy3exy2fxyy(x,y)=(2+4xy2+6xy2+4x2y4)exy2=(2+10xy2+4x2y4)exy2\begin{align*} f_x(x,y) &= y^2e^{xy^2} & f_{xx}(x,y) &= y^4e^{xy^2} & f_{xxy}(x,y) &= 4y^3e^{xy^2} + 2xy^5e^{xy^2} \\ & & f_{xy}(x,y) &= 2ye^{xy^2}+2xy^3e^{xy^2} & f_{xyy}(x,y) &= \big(2+4xy^2+6xy^2+4x^2y^4\big)e^{xy^2}\\ & & & & &= \big(2+10xy^2+4x^2y^4\big)e^{xy^2} \end{align*}

(c) We have

∂f∂u(u,v,w)=−1(u+2v+3w)2∂2f∂u ∂v(u,v,w)=4(u+2v+3w)3∂3f∂u ∂v ∂w(u,v,w)=−36(u+2v+3w)4\begin{align*} \pdiff{f}{u}(u,v,w) &= -\frac{1}{(u+2v+3w)^2} \\ \frac{\partial^2 f}{\partial u\,\partial v}(u,v,w) &= \frac{4}{(u+2v+3w)^3} \\ \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(u,v,w) &= -\frac{36}{(u+2v+3w)^4} \end{align*}

In particular

∂3f∂u ∂v ∂w(3,2,1)=−36(3+2×2+3×1)4=−36104=−92500\begin{align*} \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(3,2,1) &= -\frac{36}{(3+2\times 2+3\times 1)^4} = -\frac{36}{10^4} = -\frac{9}{2500} \end{align*}
Q4Stage 2

Find all second partial derivatives of f(x,y)=x2+5y2f(x,y)=\sqrt{x^2+5y^2}.

Answer

fxx=5y2(x2+5y2)3/2f_{xx}=\frac{5y^2}{(x^2+5y^2)^{3/2}} fxy=fyx=−5xy(x2+5y2)3/2f_{xy}=f_{yx}=-\frac{5xy}{(x^2+5y^2)^{3/2}} fyy=5x2(x2+5y2)3/2f_{yy}=\frac{5x^2}{(x^2+5y^2)^{3/2}}

Full solution

Let f(x,y)=x2+5y2f(x,y)=\sqrt{x^2+5y^2}. Then

fx=xx2+5y2fxx=1x2+5y2−12(x)(2x)(x2+5y2)3/2fxy=−12(x)(10y)(x2+5y2)3/2fy=5yx2+5y2fyy=5x2+5y2−12(5y)(10y)(x2+5y2)3/2fyx=−12(5y)(2x)(x2+5y2)3/2\begin{align*} f_x&=\frac{x}{\sqrt{x^2+5y^2}} & f_{xx}&=\frac{1}{\sqrt{x^2+5y^2}}-\frac{1}{2}\frac{(x)(2x)}{(x^2+5y^2)^{3/2}} & f_{xy}&=-\frac{1}{2}\frac{(x)(10y)}{(x^2+5y^2)^{3/2}} \cr f_y&=\frac{5y}{\sqrt{x^2+5y^2}} & f_{yy}&=\frac{5}{\sqrt{x^2+5y^2}}-\frac{1}{2}\frac{(5y)(10y)}{(x^2+5y^2)^{3/2}}& f_{yx}&=-\frac{1}{2}\frac{(5y)(2x)}{(x^2+5y^2)^{3/2}} \end{align*}

Simplifying, and in particular using that 1x2+5y2=x2+5y2(x2+5y2)3/2\frac{1}{\sqrt{x^2+5y^2}} =\frac{x^2+5y^2}{(x^2+5y^2)^{3/2}},

fxx=5y2(x2+5y2)3/2fxy=fyx=−5xy(x2+5y2)3/2fyy=5x2(x2+5y2)3/2\begin{equation*} f_{xx}=\frac{5y^2}{(x^2+5y^2)^{3/2}}\qquad f_{xy}=f_{yx}=-\frac{5xy}{(x^2+5y^2)^{3/2}}\qquad f_{yy}=\frac{5x^2}{(x^2+5y^2)^{3/2}} \end{equation*}
Q5Stage 2

Find the specified partial derivatives.

  1. f(x,y,z)=arctan⁡(exy)f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big); fxyz(x,y,z)f_{xyz}(x,y,z)

  2. f(x,y,z)=arctan⁡(exy)+arctan⁡(exz)+arctan⁡(eyz)f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big) +\arctan\big(e^{\sqrt{xz}}\big) +\arctan\big(e^{\sqrt{yz}}\big); fxyz(x,y,z)f_{xyz}(x,y,z)

  3. f(x,y,z)=arctan⁡(exyz)f(x,y,z) = \arctan\big(e^{\sqrt{xyz}}\big); fxx(1,0,0)f_{xx}(1,0,0)

Hint

(a) This higher order partial derivative can be evaluated extremely efficiently by carefully choosing the order of evaluation of the derivatives.

(b) This higher order partial derivative can be evaluated extremely efficiently by carefully choosing a different order of evaluation of the derivatives for each of the three terms.

(c) Set g(x)=f(x,0,0)g(x) = f(x,0,0). Then fxx(1,0,0)=g′′(1)f_{xx}(1,0,0)=g''(1).

Answer

(a) fxyz(x,y,z)=0f_{xyz}(x,y,z)=0 (b) fxyz(x,y,z)=0f_{xyz}(x,y,z)=0 (c) fxx(1,0,0)=0f_{xx}(1,0,0)=0

Full solution

(a) As f(x,y,z)=arctan⁡(exy)f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big) is independent of zz, we have fz(x,y,z)=0f_z(x,y,z) = 0 and hence

fxyz(x,y,z)=fzxy(x,y,z)=0\begin{equation*} f_{xyz}(x,y,z) =f_{zxy}(x,y,z) =0 \end{equation*}

(b) Write u(x,y,z)=arctan⁡(exy)u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big), v(x,y,z)=arctan⁡(exz)v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big) and w(x,y,z)=arctan⁡(eyz)w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big). Then

  • As u(x,y,z)=arctan⁡(exy)u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big) is independent of zz, we have uz(x,y,z)=0u_z(x,y,z) = 0 and hence uxyz(x,y,z)=uzxy(x,y,z)=0u_{xyz}(x,y,z) =u_{zxy}(x,y,z) =0

  • As v(x,y,z)=arctan⁡(exz)v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big) is independent of yy, we have vy(x,y,z)=0v_y(x,y,z) = 0 and hence vxyz(x,y,z)=vyxz(x,y,z)=0v_{xyz}(x,y,z) =v_{yxz}(x,y,z) =0

  • As w(x,y,z)=arctan⁡(eyz)w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big) is independent of xx, we have wx(x,y,z)=0w_x(x,y,z) = 0 and hence wxyz(x,y,z)=0w_{xyz}(x,y,z) =0

As f(x,y,z)=u(x,y,z)+v(x,y,z)+w(x,y,z)f(x,y,z)=u(x,y,z)+v(x,y,z)+w(x,y,z), we have

fxyz(x,y,z)=uxyz(x,y,z)+vxyz(x,y,z)+wxyz(x,y,z)=0\begin{equation*} f_{xyz}(x,y,z)=u_{xyz}(x,y,z)+v_{xyz}(x,y,z)+w_{xyz}(x,y,z)=0 \end{equation*}

(c) In the course of evaluating fxx(x,0,0)f_{xx}(x,0,0), both yy and zz are held fixed at 00. Thus, if we set g(x)=f(x,0,0)g(x) = f(x,0,0), then fxx(x,0,0)=g′′(x)f_{xx}(x,0,0)=g''(x). Now

g(x)=f(x,0,0)=arctan⁡(exyz)∣y=z=0=arctan⁡(1)=π4\begin{equation*} g(x) = f(x,0,0) = \arctan\big(e^{\sqrt{xyz}}\big)\Big|_{y=z=0} =\arctan(1) =\frac{\pi}{4} \end{equation*}

for all xx. So g′(x)=0g'(x)=0 and g′′(x)=0g''(x)=0 for all xx. In particular,

fxx(1,0,0)=g′′(1)=0\begin{equation*} f_{xx}(1,0,0) = g''(1) = 0 \end{equation*}
Q6Stage 2Past exam · M200 2002A

Let f(r,θ)=rmcos⁡mθf(r,\theta)=r^m\cos m\theta be a function of rr and θ\theta, where mm is a positive integer.

  1. Find the second order partial derivatives frrf_{rr}, frθf_{r\theta}, fθθf_{\theta\theta} and evaluate their respective values at (r,θ)=(1,0)(r,\theta)=(1,0).

  2. Determine the value of the real number λ\la so that f(r,θ)f(r,\theta) satisfies the differential equation

    frr+λrfr+1r2fθθ=0\begin{equation*} f_{rr}+\frac{\la}{r}f_r+\frac{1}{r^2}f_{\theta\theta}=0 \end{equation*}
Answer

(a) frr(1,0)=m(m−1), frθ(1,0)=0, fθθ(1,0)=−m2f_{rr}(1,0)=m(m-1),\ f_{r\theta}(1,0)=0,\ f_{\theta\theta}(1,0)=-m^2 (b) λ=1\la=1

Full solution

(a) The first order derivatives are

fr(r,θ)=mrm−1cos⁡mθfθ(r,θ)=−mrmsin⁡mθ\begin{equation*} f_r(r,\theta)=mr^{m-1}\cos m\theta\qquad f_\theta(r,\theta)=-mr^m\sin m\theta \end{equation*}

The second order derivatives are

frr(r,θ)=m(m−1)rm−2cos⁡mθfrθ(r,θ)=−m2rm−1sin⁡mθfθθ(r,θ)=−m2rmcos⁡mθ\begin{equation*} f_{rr}(r,\theta)=m(m-1)r^{m-2}\cos m\theta\quad f_{r\theta}(r,\theta)=-m^2r^{m-1}\sin m\theta\quad f_{\theta\theta}(r,\theta)=-m^2r^m\cos m\theta \end{equation*}

so that

frr(1,0)=m(m−1), frθ(1,0)=0, fθθ(1,0)=−m2\begin{equation*} f_{rr}(1,0)=m(m-1),\ f_{r\theta}(1,0)=0,\ f_{\theta\theta}(1,0)=-m^2 \end{equation*}

(b) By part (a), the expression

frr+λrfr+1r2fθθ=m(m−1)rm−2cos⁡mθ+λmrm−2cos⁡mθ−m2rm−2cos⁡mθ\begin{equation*} f_{rr}+\frac{\la}{r}f_r+\frac{1}{r^2}f_{\theta\theta} =m(m-1)r^{m-2}\cos m\theta+\la mr^{m-2}\cos m\theta-m^2r^{m-2}\cos m\theta \end{equation*}

vanishes for all rr and θ\theta if and only if

m(m−1)+λm−m2=0  ⟺  m(λ−1)=0  ⟺  λ=1\begin{equation*} m(m-1)+\la m-m^2=0\iff m(\la-1)=0\iff \la=1 \end{equation*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q7Stage 3

Let α>0\al>0 be a constant. Show that u(x,y,z,t)=1t3/2e−(x2+y2+z2)/(4αt)\displaystyle u(x,y,z,t) =\frac{1}{t^{3/2}} e^{-(x^2+y^2+z^2)/(4\al t)} satisfies the heat equation

ut=α(uxx+uyy+uzz)\begin{equation*} u_t = \al\big(u_{xx} + u_{yy} + u_{zz} \big) \end{equation*}

for all t>0t>0

Answer

See the solution.

Full solution

As

ut(x,y,z,t)=−321t5/2e−(x2+y2+z2)/(4αt)+14α t7/2(x2+y2+z2)e−(x2+y2+z2)/(4αt)ux(x,y,z,t)=−x2α t5/2e−(x2+y2+z2)/(4αt)uxx(x,y,z,t)=−12α t5/2e−(x2+y2+z2)/(4αt)+x24α2 t7/2e−(x2+y2+z2)/(4αt)uyy(x,y,z,t)=−12α t5/2e−(x2+y2+z2)/(4αt)+y24α2 t7/2e−(x2+y2+z2)/(4αt)uzz(x,y,z,t)=−12α t5/2e−(x2+y2+z2)/(4αt)+z24α2 t7/2e−(x2+y2+z2)/(4αt)\begin{align*} u_t(x,y,z,t) &=-\frac{3}{2}\frac{1}{t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} +\frac{1}{4\al\,t^{7/2}}(x^2+y^2+z^2) e^{-(x^2+y^2+z^2)/(4\al t)} \\ u_x(x,y,z,t) &=-\frac{x}{2\al\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} \\ u_{xx}(x,y,z,t) &=-\frac{1}{2\al\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} +\frac{x^2}{4\al^2\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\al t)} \\ u_{yy}(x,y,z,t) &=-\frac{1}{2\al\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} +\frac{y^2}{4\al^2\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\al t)} \\ u_{zz}(x,y,z,t) &=-\frac{1}{2\al\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} +\frac{z^2}{4\al^2\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\al t)} \end{align*}

we have

α(uxx+uyy+uzz)=−32 t5/2e−(x2+y2+z2)/(4αt)+x2+y2+z24α t7/2e−(x2+y2+z2)/(4αt)=ut\begin{align*} \al\big(u_{xx} + u_{yy} + u_{zz} \big) &=-\frac{3}{2\,t^{5/2}} e^{-(x^2+y^2+z^2)/(4\al t)} +\frac{x^2+y^2+z^2}{4\al\,t^{7/2}} e^{-(x^2+y^2+z^2)/(4\al t)} =u_t \end{align*}

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From the UBC Math 200 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.