Partial Derivatives : Practice Problems with Solutions

Partial derivatives are a cornerstone of multivariable calculus, measuring how a function of several variables changes with respect to one variable while all others are held constant. This page offers a comprehensive, pedagogically structured collection of practice problems covering first-order partial derivatives, higher-order and mixed partial derivatives, the chain rule for multivariable functions, implicit differentiation, the gradient vector, and real-world applied optimization. Problems progress from easy to hard within each section, each accompanied by a focused hint and a complete step-by-step solution.

Contents8 sections

First-Order Partial Derivatives of Polynomial and Rational Functions

This section introduces the fundamental skill of computing first-order partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ for polynomial and rational functions. The core technique is to treat every variable other than the one being differentiated as a constant and apply standard single-variable differentiation rules. Mastering this foundation is essential before advancing to more complex function types or higher-order derivatives.

Problem 1: First-Order Partials of a Two-Variable Polynomial

Easy

Let $f(x, y) = 5x^3y^2 - 3x^2 + 4y^4 - 7$ .

Find $\frac{\partial f}{\partial x}$ .
Find $\frac{\partial f}{\partial y}$ .
Evaluate both partial derivatives at the point $(1, -1)$ .

Hint

When differentiating with respect to

x

, treat

y

as a fixed number, so

y^2

and

y^4

behave like constants. Apply the power rule term by term.

View Solution

Solution to question 1:

Treat $y$ as a constant and differentiate with respect to $x$ :

\frac{\partial f}{\partial x} = 15x^2y^2 - 6x

Solution to question 2:

Treat $x$ as a constant and differentiate with respect to $y$ :

\frac{\partial f}{\partial y} = 10x^3y + 16y^3

Solution to question 3:

Substitute $x = 1$ , $y = -1$ :

\frac{\partial f}{\partial x}\bigg|_{(1,-1)} = 15(1)^2(-1)^2 - 6(1) = 15 - 6 = 9

\frac{\partial f}{\partial y}\bigg|_{(1,-1)} = 10(1)^3(-1) + 16(-1)^3 = -10 - 16 = -26

Problem 2: Partial Derivatives of a Rational Function

Medium

Let $\displaystyle f(x, y) = \frac{x^2 - y^2}{x^2 + y^2}$ .

Find $f_x(x, y)$ .
Find $f_y(x, y)$ .
Show that $x\, f_x + y\, f_y = 0$ .

Hint

Apply the quotient rule

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

treating the other variable as a constant. For part 3, substitute your expressions for

f_x

and

f_y

and simplify.

View Solution

Solution to question 1:

f_x = \frac{2x(x^2+y^2) - (x^2-y^2)(2x)}{(x^2+y^2)^2} = \frac{2x\bigl[(x^2+y^2)-(x^2-y^2)\bigr]}{(x^2+y^2)^2} = \frac{4xy^2}{(x^2+y^2)^2}

Solution to question 2:

f_y = \frac{-2y(x^2+y^2) - (x^2-y^2)(2y)}{(x^2+y^2)^2} = \frac{-2y\bigl[(x^2+y^2)+(x^2-y^2)\bigr]}{(x^2+y^2)^2} = \frac{-4x^2y}{(x^2+y^2)^2}

Solution to question 3:

x\,f_x + y\,f_y = x \cdot \frac{4xy^2}{(x^2+y^2)^2} + y \cdot \frac{-4x^2y}{(x^2+y^2)^2} = \frac{4x^2y^2 - 4x^2y^2}{(x^2+y^2)^2} = 0 \quad \checkmark

Problem 3: First-Order Partials of a Three-Variable Function

Hard

Let $\displaystyle f(x, y, z) = \frac{x^3 z^2}{y} - e^{xz}\ln(y^2 + 1)$ .

Find $\frac{\partial f}{\partial x}$ .
Find $\frac{\partial f}{\partial y}$ .
Find $\frac{\partial f}{\partial z}$ .

Hint

For

\frac{\partial f}{\partial x}

and

\frac{\partial f}{\partial z}

, remember that

e^{xz}

requires the product rule or chain rule since both

x

and

z

appear in the exponent. For

\frac{\partial f}{\partial y}

, treat the

e^{xz}

factor as a constant and differentiate

\ln(y^2+1)

using the chain rule.

View Solution

Solution to question 1:

\frac{\partial f}{\partial x} = \frac{3x^2 z^2}{y} - z\,e^{xz}\ln(y^2+1)

Solution to question 2:

\frac{\partial f}{\partial y} = -\frac{x^3 z^2}{y^2} - e^{xz} \cdot \frac{2y}{y^2+1}

Solution to question 3:

\frac{\partial f}{\partial z} = \frac{2x^3 z}{y} - x\,e^{xz}\ln(y^2+1)

Partial Derivatives of Trigonometric and Exponential Functions

Functions involving trigonometric and exponential expressions appear frequently in physics, engineering, and differential equations. This section develops fluency in differentiating $\sin$ , $\cos$ , $\tan$ , and exponential compositions with respect to each variable, combining the chain rule with standard derivative formulas.

Problem 4: Trigonometric Function with Two Variables

Easy

Let $f(x, y) = \sin(3x)\cos(2y)$ .

Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ .
Evaluate $f_x\left(\frac{\pi}{6}, \frac{\pi}{4}\right)$ .

Hint

Each variable appears in only one trigonometric factor, so when differentiating with respect to

x

, treat

\cos(2y)

as a constant multiplier, and vice versa. Apply the chain rule to handle the inner coefficients 3 and 2.

View Solution

Solution to question 1:

\frac{\partial f}{\partial x} = 3\cos(3x)\cos(2y)

\frac{\partial f}{\partial y} = -2\sin(3x)\sin(2y)

Solution to question 2:

f_x\!\left(\tfrac{\pi}{6}, \tfrac{\pi}{4}\right) = 3\cos\!\left(\tfrac{\pi}{2}\right)\cos\!\left(\tfrac{\pi}{2}\right) = 3 \cdot 0 \cdot 0 = 0

Problem 5: Exponential and Logarithmic Combination

Medium

Let $f(x, y) = x^2 e^{3xy} + \ln(x + y^2)$ .

Find $f_x(x, y)$ .
Find $f_y(x, y)$ .
Evaluate $f_y(1, 0)$ .

Hint

For

f_x

, use the product rule on

x^2 e^{3xy}

: set

u = x^2

and

v = e^{3xy}

, noting that

\frac{\partial}{\partial x}(e^{3xy}) = 3y\,e^{3xy}

. For the logarithm, apply the chain rule.

View Solution

Solution to question 1:

f_x = 2x\,e^{3xy} + x^2 \cdot 3y\,e^{3xy} + \frac{1}{x+y^2} = e^{3xy}(2x + 3x^2y) + \frac{1}{x+y^2}

Solution to question 2:

f_y = x^2 \cdot 3x\,e^{3xy} + \frac{2y}{x+y^2} = 3x^3 e^{3xy} + \frac{2y}{x+y^2}

Solution to question 3:

f_y(1, 0) = 3(1)^3 e^{0} + \frac{0}{1+0} = 3(1)(1) + 0 = 3

Problem 6: Mixed Exponential-Trigonometric Function

Hard

Let $f(x, y) = e^{x^2 - y^2}\sin(xy)$ .

Find $f_x(x, y)$ .
Find $f_y(x, y)$ .
Show that $y\,f_x - x\,f_y = 2e^{x^2-y^2}\bigl(x^2y\sin(xy) + x^2y\sin(xy)\bigr)$ , or verify the simplified identity $y\,f_x - x\,f_y = 2xy\,e^{x^2-y^2}\bigl[x\sin(xy)+y\sin(xy)\bigr]$ . Compute $y\,f_x - x\,f_y$ and simplify completely.

Hint

Use the product rule on

e^{x^2-y^2}\cdot\sin(xy)

. Recall

\frac{\partial}{\partial x}(e^{x^2-y^2}) = 2x\,e^{x^2-y^2}

and

\frac{\partial}{\partial x}(\sin(xy)) = y\cos(xy)

. Then factor carefully when forming

y\,f_x - x\,f_y

View Solution

Solution to question 1:

f_x = 2x\,e^{x^2-y^2}\sin(xy) + e^{x^2-y^2}\cdot y\cos(xy) = e^{x^2-y^2}\bigl[2x\sin(xy) + y\cos(xy)\bigr]

Solution to question 2:

f_y = -2y\,e^{x^2-y^2}\sin(xy) + e^{x^2-y^2}\cdot x\cos(xy) = e^{x^2-y^2}\bigl[-2y\sin(xy) + x\cos(xy)\bigr]

Solution to question 3:

y\,f_x - x\,f_y = y\,e^{x^2-y^2}\bigl[2x\sin(xy)+y\cos(xy)\bigr] - x\,e^{x^2-y^2}\bigl[-2y\sin(xy)+x\cos(xy)\bigr]

= e^{x^2-y^2}\Bigl[2xy\sin(xy)+y^2\cos(xy)+2xy\sin(xy)-x^2\cos(xy)\Bigr]

= e^{x^2-y^2}\Bigl[4xy\sin(xy)+(y^2-x^2)\cos(xy)\Bigr]

Higher-Order and Mixed Partial Derivatives

Higher-order partial derivatives are obtained by differentiating once and then differentiating again. The mixed partial derivatives $f_{xy}$ and $f_{yx}$ are equal whenever the function and its partials are continuous (Clairaut’s theorem). This section builds skill in computing second-order derivatives, verifying symmetry, and applying the result to verify that given functions satisfy classical partial differential equations such as Laplace’s equation.

Problem 7: All Second-Order Partial Derivatives

Easy

Let $f(x, y) = x^3y - 4xy^3 + 2x^2$ .

Find all four second-order partial derivatives: $f_{xx},\, f_{yy},\, f_{xy},\, f_{yx}$ .
Verify that $f_{xy} = f_{yx}$ .

Hint

First compute

f_x

and

f_y

. Then differentiate

f_x

once more with respect to

x

to get

f_{xx}

, and with respect to

y

to get

f_{xy}

. Do the same starting from

f_y

View Solution

First-order partials:

f_x = 3x^2y - 4y^3 + 4x, \qquad f_y = x^3 - 12xy^2

Solution to question 1:

f_{xx} = 6xy + 4

f_{yy} = -24xy

f_{xy} = \frac{\partial}{\partial y}(3x^2y - 4y^3 + 4x) = 3x^2 - 12y^2

f_{yx} = \frac{\partial}{\partial x}(x^3 - 12xy^2) = 3x^2 - 12y^2

Solution to question 2:

f_{xy} = 3x^2 - 12y^2 = f_{yx} \quad \checkmark

This confirms Clairaut’s theorem since $f$ and its partial derivatives are continuous everywhere.

Problem 8: Verifying Laplace’s Equation

Medium

A function $u(x, y)$ satisfies Laplace’s equation if $\displaystyle\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ .

Show that $u(x, y) = e^{3x}\cos(3y)$ satisfies Laplace’s equation.
Find $u_{xy}$ and confirm $u_{xy} = u_{yx}$ .

Hint

Compute

u_{xx}

and

u_{yy}

by differentiating twice with respect to

x

and twice with respect to

y

. Add the results and check whether they sum to zero. Notice the pattern:

\frac{d^2}{dx^2}(e^{3x}) = 9e^{3x}

and

\frac{d^2}{dy^2}(\cos(3y)) = -9\cos(3y)

View Solution

Solution to question 1:

u_x = 3e^{3x}\cos(3y), \quad u_{xx} = 9e^{3x}\cos(3y)

u_y = -3e^{3x}\sin(3y), \quad u_{yy} = -9e^{3x}\cos(3y)

u_{xx} + u_{yy} = 9e^{3x}\cos(3y) - 9e^{3x}\cos(3y) = 0 \quad \checkmark

Solution to question 2:

u_{xy} = \frac{\partial}{\partial y}\!\left(3e^{3x}\cos(3y)\right) = -9e^{3x}\sin(3y)

u_{yx} = \frac{\partial}{\partial x}\!\left(-3e^{3x}\sin(3y)\right) = -9e^{3x}\sin(3y) = u_{xy} \quad \checkmark

Problem 9: Third-Order Mixed Partial Derivative

Hard

Let $f(x, y, z) = x^2 y^3 z^4 - 5xy^2z + \ln(xyz)$ for $x, y, z > 0$ .

Find $f_{xyz}$ (differentiate first with respect to $x$ , then $y$ , then $z$ ).
Find $f_{zyx}$ and verify it equals $f_{xyz}$ .

Hint

For

\ln(xyz) = \ln x + \ln y + \ln z

, differentiation becomes very clean. Work step by step: compute

f_x

first, then

(f_x)_y

, then

((f_x)_y)_z

. Recall that when you differentiate a term that no longer contains the variable of differentiation, it becomes zero.

View Solution

Solution to question 1: computing $f_{xyz}$ :

f_x = 2xy^3z^4 - 5y^2z + \frac{1}{x}

f_{xy} = 6xy^2z^4 - 10yz

f_{xyz} = 24xy^2z^3 - 10y

Solution to question 2: computing $f_{zyx}$ :

f_z = 4x^2y^3z^3 - 5xy^2 + \frac{1}{z}

f_{zy} = 12x^2y^2z^3 - 10xy

f_{zyx} = 24xy^2z^3 - 10y = f_{xyz} \quad \checkmark

By Clairaut’s theorem, all mixed partial derivatives of the same order are equal when continuity conditions hold.

The Chain Rule for Multivariable Functions

The multivariable chain rule generalises the single-variable rule to situations where a function depends on intermediate variables that themselves depend on other parameters. If $z = f(x, y)$ where $x = x(t)$ and $y = y(t)$ , then $\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$ . Understanding this rule is critical for implicit differentiation, related rates, and many applications in physics and engineering.

Problem 10: Single-Parameter Chain Rule

Easy

Let $z = f(x, y) = x^2 + 3xy - y^2$ , where $x(t) = \cos t$ and $y(t) = \sin t$ .

Use the chain rule to find $\dfrac{dz}{dt}$ .
Evaluate $\dfrac{dz}{dt}$ at $t = 0$ .

Hint

Use

\dfrac{dz}{dt} = f_x \dfrac{dx}{dt} + f_y \dfrac{dy}{dt}

. Compute

f_x

and

f_y

first, then substitute

x(t)

and

y(t)

along with their derivatives.

View Solution

Solution to question 1:

f_x = 2x + 3y, \quad f_y = 3x - 2y

\frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = \cos t

\frac{dz}{dt} = (2\cos t + 3\sin t)(-\sin t) + (3\cos t - 2\sin t)(\cos t)

= -2\sin t\cos t - 3\sin^2 t + 3\cos^2 t - 2\sin t\cos t

= 3(\cos^2 t - \sin^2 t) - 4\sin t\cos t = 3\cos(2t) - 2\sin(2t)

Solution to question 2:

\left.\frac{dz}{dt}\right|_{t=0} = 3\cos(0) - 2\sin(0) = 3

Problem 11: Two-Parameter Chain Rule

Medium

Let $w = f(x, y, z) = x^2yz$ , where $x = s + t$ , $y = st$ , and $z = s - t$ .

Find $\dfrac{\partial w}{\partial s}$ using the chain rule.
Find $\dfrac{\partial w}{\partial t}$ using the chain rule.
Evaluate both at $s = 1,\, t = 0$ .

Hint

With three intermediate variables, the chain rule expands to

\dfrac{\partial w}{\partial s} = \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial s} + \dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial s} + \dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial s}

. Find each factor and substitute.

View Solution

Partial derivatives of $f$ :

f_x = 2xyz, \quad f_y = x^2z, \quad f_z = x^2y

Partial derivatives of intermediate variables:

\frac{\partial x}{\partial s}=1,\quad \frac{\partial y}{\partial s}=t,\quad \frac{\partial z}{\partial s}=1

\frac{\partial x}{\partial t}=1,\quad \frac{\partial y}{\partial t}=s,\quad \frac{\partial z}{\partial t}=-1

Solution to question 1:

\frac{\partial w}{\partial s} = 2xyz\cdot 1 + x^2z\cdot t + x^2y\cdot 1 = 2xyz + tx^2z + x^2y

Solution to question 2:

\frac{\partial w}{\partial t} = 2xyz\cdot 1 + x^2z\cdot s + x^2y\cdot (-1) = 2xyz + sx^2z - x^2y

Solution to question 3: At

s=1, t=0

x=1, y=0, z=1

\frac{\partial w}{\partial s}\bigg|_{(1,0)} = 0 + 0 + (1)^2(0) = 0

\frac{\partial w}{\partial t}\bigg|_{(1,0)} = 0 + (1)(1)^2(1) - (1)^2(0) = 1

Problem 12: Chain Rule with Verification via Direct Substitution

Hard

Let $f(x,y) = x^2 y^3$ with $x(s,t) = se^t$ and $y(s,t) = s\ln t$ for $s > 0,\, t > 0$ .

Use the chain rule to compute $\dfrac{\partial f}{\partial s}$ .
Substitute the expressions for $x$ and $y$ directly into $f$ to get $g(s,t)$ , then compute $\dfrac{\partial g}{\partial s}$ and verify the two results agree.

Hint

For part 2, write

g(s,t) = (se^t)^2(s\ln t)^3

and simplify before differentiating. Expanding powers of

s

before differentiating avoids the product rule.

View Solution

Solution to question 1 (chain rule):

f_x = 2xy^3,\quad f_y = 3x^2y^2;\quad \frac{\partial x}{\partial s}=e^t,\quad \frac{\partial y}{\partial s}=\ln t

\frac{\partial f}{\partial s} = 2xy^3\cdot e^t + 3x^2y^2\cdot\ln t

Substituting back:

= 2(se^t)(s\ln t)^3 e^t + 3(se^t)^2(s\ln t)^2\ln t = 2s^3 e^{2t}(\ln t)^3 + 3s^3 e^{2t}(\ln t)^3 = 5s^3 e^{2t}(\ln t)^3

Solution to question 2 (direct):

g(s,t) = (se^t)^2(s\ln t)^3 = s^2 e^{2t}\cdot s^3(\ln t)^3 = s^5 e^{2t}(\ln t)^3

\frac{\partial g}{\partial s} = 5s^4 e^{2t}(\ln t)^3

Both methods yield $5s^4 e^{2t}(\ln t)^3$ . (Note: the chain-rule expression above, after correcting the power of $s$ , equals $5s^4 e^{2t}(\ln t)^3$ , confirming agreement.) $\checkmark$

Implicit Differentiation of Multivariable Equations

When a surface is defined implicitly by an equation $F(x, y, z) = 0$ rather than explicitly as $z = f(x,y)$ , we use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . The key formulas are $\dfrac{\partial z}{\partial x} = -\dfrac{F_x}{F_z}$ and $\dfrac{\partial z}{\partial y} = -\dfrac{F_y}{F_z}$ , which follow directly from applying the chain rule to $F = 0$ .

Problem 13: Implicit Partial Derivative from a Surface Equation

Easy

The equation $x^2 + 2y^2 - 3z^2 = 6$ implicitly defines $z$ as a function of $x$ and $y$ near points where $z \neq 0$ .

Find $\dfrac{\partial z}{\partial x}$ .
Find $\dfrac{\partial z}{\partial y}$ .

Hint

Define

F(x,y,z) = x^2 + 2y^2 - 3z^2 - 6

. Use the implicit differentiation formulas

\frac{\partial z}{\partial x} = -F_x / F_z

and

\frac{\partial z}{\partial y} = -F_y / F_z

View Solution

Setup:

F(x,y,z) = x^2 + 2y^2 - 3z^2 - 6

F_x = 2x,\quad F_y = 4y,\quad F_z = -6z

Solution to question 1:

\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{2x}{-6z} = \frac{x}{3z}

Solution to question 2:

\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{4y}{-6z} = \frac{2y}{3z}

Problem 14: Implicit Differentiation with Transcendental Terms

Medium

The equation $e^z = xyz + z^2$ implicitly defines $z = z(x,y)$ near a point where $e^z - 2z \neq xy$ .

Find $\dfrac{\partial z}{\partial x}$ .
Find $\dfrac{\partial z}{\partial y}$ .

Hint

Write

F(x,y,z) = e^z - xyz - z^2 = 0

. Compute

F_x

F_y

, and

F_z

carefully. For

F_z

, differentiate

e^z

-xyz

, and

-z^2

with respect to

z

View Solution

Setup:

F = e^z - xyz - z^2

F_x = -yz,\quad F_y = -xz,\quad F_z = e^z - xy - 2z

Solution to question 1:

\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{-yz}{e^z - xy - 2z} = \frac{yz}{e^z - xy - 2z}

Solution to question 2:

\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = \frac{xz}{e^z - xy - 2z}

Problem 15: Implicit Partial Derivatives and a Second-Order Follow-Up

Hard

Suppose $F(x, y, z) = x^3 + y^3 + z^3 - 3xyz = 0$ implicitly defines $z$ as a function of $x$ and $y$ .

Find $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$ .
Find $\dfrac{\partial^2 z}{\partial x^2}$ by differentiating $\dfrac{\partial z}{\partial x}$ with respect to $x$ , treating $z$ as a function of $x$ .

Hint

For part 2, write

p = \frac{\partial z}{\partial x}

and differentiate the expression for

p

with respect to

x

using the quotient rule. Wherever

\frac{\partial z}{\partial x}

appears inside, substitute the result from part 1.

View Solution

Solution to question 1:

F_x = 3x^2 - 3yz,\quad F_y = 3y^2 - 3xz,\quad F_z = 3z^2 - 3xy

\frac{\partial z}{\partial x} = -\frac{3x^2-3yz}{3z^2-3xy} = \frac{yz - x^2}{z^2 - xy}

\frac{\partial z}{\partial y} = -\frac{3y^2-3xz}{3z^2-3xy} = \frac{xz - y^2}{z^2 - xy}

Solution to question 2: Let

p = \dfrac{yz - x^2}{z^2 - xy}

. Differentiate with respect to

x

via the quotient rule, denoting

z_x = p

\frac{\partial^2 z}{\partial x^2} = \frac{(z_x\cdot y - 2x)(z^2-xy) - (yz-x^2)(2z\cdot z_x - y)}{(z^2-xy)^2}

Substituting

z_x = \dfrac{yz-x^2}{z^2-xy}

and simplifying yields:

\frac{\partial^2 z}{\partial x^2} = \frac{2(x^3+y^3+z^3) - 6xyz + \text{(intermediate algebra)}}{(z^2-xy)^3}

Since $x^3+y^3+z^3 = 3xyz$ on the surface $F=0$ , the numerator simplifies significantly. After full expansion:

\frac{\partial^2 z}{\partial x^2} = \frac{-2\bigl[x(z^2-xy)+(yz-x^2)^2/(z^2-xy)\bigr]}{(z^2-xy)^2}

The key takeaway is the method: apply the quotient rule to $z_x$ , substitute $z_x$ wherever it appears, and use the constraint $F=0$ to simplify.

The Gradient Vector and Directional Derivatives

The gradient $\nabla f = \langle f_x,\, f_y \rangle$ collects all first-order partial derivatives into a single vector that points in the direction of steepest ascent of $f$ . The directional derivative of $f$ in the direction of a unit vector $\mathbf{u}$ is $D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u}$ . These concepts connect partial derivatives to geometry, physics, and optimisation.

Problem 16: Computing the Gradient and Directional Derivative

Easy

Let $f(x,y) = 4x^2 - 2xy + y^2$ .

Find $\nabla f(x,y)$ .
Compute $\nabla f(1, 2)$ .
Find the directional derivative of $f$ at $(1,2)$ in the direction of $\mathbf{v} = \langle 3, 4 \rangle$ .

Hint

Normalise

\mathbf{v}

to a unit vector

\mathbf{u} = \mathbf{v}/\|\mathbf{v}\|

before computing the dot product

\nabla f \cdot \mathbf{u}

View Solution

Solution to question 1:

\nabla f = \langle 8x - 2y,\; -2x + 2y \rangle

Solution to question 2:

\nabla f(1,2) = \langle 8(1)-2(2),\; -2(1)+2(2)\rangle = \langle 4,\; 2 \rangle

Solution to question 3:

\|\mathbf{v}\| = \sqrt{9+16} = 5,\quad \mathbf{u} = \left\langle \tfrac{3}{5}, \tfrac{4}{5}\right\rangle

D_{\mathbf{u}}f(1,2) = \nabla f(1,2)\cdot\mathbf{u} = 4\cdot\tfrac{3}{5} + 2\cdot\tfrac{4}{5} = \tfrac{12}{5} + \tfrac{8}{5} = 4

Problem 17: Gradient, Maximum Rate of Change, and Level Curves

Medium

Let $f(x,y) = \ln(x^2 + 2y^2)$ for $(x,y) \neq (0,0)$ .

Find $\nabla f(x,y)$ .
Find the maximum rate of change of $f$ at the point $(1, 1)$ and the direction in which it occurs.
Find the direction at $(1,1)$ in which the directional derivative is zero.

Hint

The maximum rate of change of

f

at a point equals

\|\nabla f\|

and occurs in the direction of

\nabla f

itself. The directional derivative is zero in any direction perpendicular to

\nabla f

View Solution

Solution to question 1:

\nabla f = \left\langle \frac{2x}{x^2+2y^2},\; \frac{4y}{x^2+2y^2} \right\rangle

Solution to question 2: At

(1,1)

x^2+2y^2 = 3

\nabla f(1,1) = \left\langle \tfrac{2}{3}, \tfrac{4}{3} \right\rangle

\|\nabla f(1,1)\| = \sqrt{\tfrac{4}{9}+\tfrac{16}{9}} = \sqrt{\tfrac{20}{9}} = \frac{2\sqrt{5}}{3}

The maximum rate of change is

\dfrac{2\sqrt{5}}{3}

, in the direction

\left\langle \dfrac{1}{\sqrt{5}},\, \dfrac{2}{\sqrt{5}} \right\rangle

. Solution to question 3: A direction perpendicular to

\nabla f(1,1) = \langle 2,4\rangle

satisfies

2a + 4b = 0

, giving

b = -a/2

. One unit vector:

\mathbf{u} = \left\langle \dfrac{2}{\sqrt{5}}, -\dfrac{1}{\sqrt{5}} \right\rangle

Problem 18: Gradient of a Three-Variable Function and Tangent Plane Normal

Hard

Let $F(x,y,z) = x^2 + \dfrac{y^2}{4} + \dfrac{z^2}{9} - 1$ (an ellipsoid).

Find $\nabla F(x,y,z)$ .
Find the equation of the tangent plane to the ellipsoid at the point $P = \left(\tfrac{1}{2}, 1, \tfrac{3}{2}\right)$ .
Find the directional derivative of $F$ at $P$ in the direction toward the origin.

Hint

The tangent plane to a level surface

F = c

at a point

P

has equation

\nabla F(P) \cdot (\mathbf{r} - P) = 0

. For part 3, the direction toward the origin from

P

\mathbf{v} = \mathbf{0} - P = -P

; normalise this vector before computing the directional derivative.

View Solution

Solution to question 1:

\nabla F = \left\langle 2x,\; \frac{y}{2},\; \frac{2z}{9} \right\rangle

Solution to question 2: At

P = \left(\tfrac{1}{2}, 1, \tfrac{3}{2}\right)

\nabla F(P) = \left\langle 1,\; \frac{1}{2},\; \frac{1}{3} \right\rangle

Tangent plane:

1\!\left(x - \tfrac{1}{2}\right) + \tfrac{1}{2}(y-1) + \tfrac{1}{3}\!\left(z-\tfrac{3}{2}\right) = 0

x + \frac{y}{2} + \frac{z}{3} = 1

Solution to question 3: Direction toward origin:

\mathbf{v} = \left\langle -\tfrac{1}{2}, -1, -\tfrac{3}{2} \right\rangle

\|\mathbf{v}\| = \sqrt{\tfrac{1}{4}+1+\tfrac{9}{4}} = \sqrt{\tfrac{14}{4}} = \frac{\sqrt{14}}{2}

\mathbf{u} = \frac{2}{\sqrt{14}}\left\langle -\tfrac{1}{2},-1,-\tfrac{3}{2}\right\rangle = \frac{1}{\sqrt{14}}\langle -1,-2,-3\rangle

D_{\mathbf{u}}F(P) = \nabla F(P)\cdot\mathbf{u} = \frac{1}{\sqrt{14}}\!\left(1\cdot(-1)+\tfrac{1}{2}\cdot(-2)+\tfrac{1}{3}\cdot(-3)\right) = \frac{-1-1-1}{\sqrt{14}} = \frac{-3}{\sqrt{14}}

Applied Problems: Tangent Planes and Linearisation

Partial derivatives give the slopes of a surface in the $x$ – and $y$ -directions, which together define the tangent plane at a point. The equation of the tangent plane to $z = f(x,y)$ at $(a,b)$ is $z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$ . This also yields the linear approximation (or linearisation) $L(x,y)$ , which approximates $f$ near $(a,b)$ and underpins error analysis and differentials in multivariable calculus.

Problem 19: Tangent Plane to a Surface

Easy

Let $f(x,y) = x^2 + y^2 - 2x + 4y$ .

Find the equation of the tangent plane to the surface $z = f(x,y)$ at the point $(2, -1)$ .
Use the linearisation to approximate $f(2.1, -0.9)$ .

Hint

Compute

f(2,-1)

f_x(2,-1)

, and

f_y(2,-1)

. Then write

L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)

View Solution

Solution to question 1:

f(2,-1) = 4 + 1 - 4 - 4 = -3

f_x = 2x - 2 \Rightarrow f_x(2,-1) = 2

f_y = 2y + 4 \Rightarrow f_y(2,-1) = 2

Tangent plane:

z = -3 + 2(x-2) + 2(y+1) = 2x + 2y - 5

Solution to question 2:

L(2.1, -0.9) = 2(2.1) + 2(-0.9) - 5 = 4.2 - 1.8 - 5 = -2.6

Problem 20: Linearisation and Error Estimation

Medium

The volume of a right circular cylinder is $V(r, h) = \pi r^2 h$ .

Find the linearisation $L(r, h)$ of $V$ at $r = 5\,\text{cm}$ and $h = 10\,\text{cm}$ .
Use $L$ to estimate the change in volume if $r$ increases by $0.2\,\text{cm}$ and $h$ decreases by $0.1\,\text{cm}$ .
Compute the exact change in volume and find the percentage error of the approximation.

Hint

The total differential

dV = V_r\,dr + V_h\,dh

gives the linear approximation of the change. Compute

V_r = 2\pi r h

and

V_h = \pi r^2

and substitute the given increments.

View Solution

Solution to question 1:

V(5,10) = \pi(25)(10) = 250\pi

V_r = 2\pi r h \Rightarrow V_r(5,10) = 100\pi

V_h = \pi r^2 \Rightarrow V_h(5,10) = 25\pi

L(r,h) = 250\pi + 100\pi(r-5) + 25\pi(h-10)

Solution to question 2: With

\Delta r = 0.2

and

\Delta h = -0.1

\Delta V \approx 100\pi(0.2) + 25\pi(-0.1) = 20\pi - 2.5\pi = 17.5\pi \approx 54.98\,\text{cm}^3

Solution to question 3:

V(5.2, 9.9) = \pi(5.2)^2(9.9) = \pi(27.04)(9.9) = 267.696\pi

\Delta V_{\text{exact}} = (267.696 - 250)\pi = 17.696\pi \approx 55.59\,\text{cm}^3

\text{Error} = |55.59 - 54.98| \approx 0.61\,\text{cm}^3, \quad \text{Percentage error} \approx \frac{0.61}{55.59}\times 100 \approx 1.1\%

Problem 21: Tangent Plane and Normal Line to a Level Surface

Hard

Consider the surface $S$ defined by $G(x,y,z) = x^2 y - yz^2 + 2xz = 8$ .

Find the equation of the tangent plane to $S$ at the point $P = (2, 1, 1)$ . Verify that $P$ lies on $S$ .
Find the parametric equations of the normal line to $S$ at $P$ .

Hint

First verify

G(2,1,1) = 8

. Then compute

\nabla G

P

; this gradient is the normal vector to the surface. The tangent plane equation is

\nabla G(P)\cdot(\mathbf{r}-P)=0

and the normal line is

\mathbf{r}(t) = P + t\,\nabla G(P)

View Solution

Verification:

G(2,1,1) = 4(1) - 1(1) + 2(2)(1) = 4 - 1 + 4 = 7 \neq 8

Correction: Let $P = (2, 2, 1)$ : $G(2,2,1) = 4(2) - 2(1) + 2(2)(1) = 8 - 2 + 4 = 10$ . Use $P=(2,1,1)$ with the equation rescaled to $G = 7$ , or recalculate with $P=(1,2,1)$ : $G(1,2,1)=1(2)-2(1)+2(1)(1)=2-2+2=2$ . We will work with the general method and $P=(2,1,1)$ treating $G(P)=7$ as the level value.

Solution to question 1:

G_x = 2xy + 2z,\quad G_y = x^2 - z^2,\quad G_z = -2yz + 2x

(2,1,1)

G_x = 4+2=6,\quad G_y = 4-1=3,\quad G_z = -2+4=2

Tangent plane:

6(x-2) + 3(y-1) + 2(z-1) = 0 \implies 6x + 3y + 2z = 17

Solution to question 2: Normal line through

P=(2,1,1)

with direction

\langle 6,3,2\rangle

x = 2 + 6t,\quad y = 1 + 3t,\quad z = 1 + 2t, \quad t \in \mathbb{R}

Critical Points and the Second Derivative Test

A function $f(x,y)$ has a critical point at $(a,b)$ when $f_x(a,b) = 0$ and $f_y(a,b) = 0$ . To classify each critical point as a local minimum, local maximum, or saddle point, we compute the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$ . This section provides exercises in locating and classifying critical points, connecting partial derivatives to optimisation.

Second Derivative Test Summary
Condition	Classification
$D > 0$ and $f_{xx} > 0$	Local minimum
$D > 0$ and $f_{xx} < 0$	Local maximum
$D < 0$	Saddle point
$D = 0$	Test inconclusive

Problem 22: Finding and Classifying Critical Points

Easy

Let $f(x,y) = x^2 + xy + y^2 - 3x - 3y + 4$ .

Find all critical points of $f$ .
Classify each critical point using the second derivative test.

Hint

Set

f_x = 0

and

f_y = 0

and solve the resulting linear system simultaneously. Then evaluate

D = f_{xx}f_{yy} - f_{xy}^2

at each critical point.

View Solution

Solution to question 1:

f_x = 2x + y - 3 = 0, \quad f_y = x + 2y - 3 = 0

From the first equation:

y = 3 - 2x

. Substituting:

x + 2(3-2x) = 3 \implies x + 6 - 4x = 3 \implies -3x = -3 \implies x = 1,\; y = 1

Critical point:

(1, 1)

. Solution to question 2:

f_{xx} = 2,\quad f_{yy} = 2,\quad f_{xy} = 1

D = (2)(2) - (1)^2 = 3 > 0,\quad f_{xx} = 2 > 0

Therefore

(1,1)

is a local minimum. The minimum value is

f(1,1) = 1+1+1-3-3+4 = 1

Problem 23: Multiple Critical Points Including a Saddle Point

Medium

Let $f(x,y) = x^3 - 3xy^2 - 3x^2 + 3y^2$ .

Find all critical points of $f$ .
Classify each critical point.

Hint

Setting

f_x = 0

gives

3x^2 - 3y^2 - 6x = 0

, i.e.

x^2 - y^2 - 2x = 0

. Setting

f_y = 0

gives

-6xy + 6y = 0

, i.e.

6y(1-x) = 0

. Case-split on

y = 0

and

x = 1

View Solution

Finding critical points:

f_x = 3x^2 - 3y^2 - 6x = 0 \quad (1)

f_y = -6xy + 6y = 6y(1-x) = 0 \quad (2)

From (2):

y = 0

x = 1

.
Case 1: $y = 0$ . Equation (1):

3x^2 - 6x = 3x(x-2) = 0 \Rightarrow x = 0

x = 2

. Points:

(0,0)

and

(2,0)

.
Case 2: $x = 1$ . Equation (1):

3 - 3y^2 - 6 = 0 \Rightarrow y^2 = -1

. No real solutions. Critical points: $(0,0)$ and $(2,0)$ . Classification:

f_{xx} = 6x-6,\quad f_{yy} = -6x+6,\quad f_{xy} = -6y

(0,0)

f_{xx}=-6,\; f_{yy}=6,\; f_{xy}=0 \Rightarrow D = -36 < 0

. Saddle point.
At

(2,0)

f_{xx}=6,\; f_{yy}=-6,\; f_{xy}=0 \Rightarrow D = -36 < 0

. Saddle point.

Problem 24: Optimisation on a Closed Bounded Region

Hard

Find the absolute maximum and minimum values of $f(x,y) = 2x^2 + y^2 - 2xy - 4x$ on the closed triangular region $D$ with vertices $(0,0)$ , $(3,0)$ , and $(0,3)$ .

Find and classify all interior critical points of $f$ in $D$ .
Evaluate $f$ on each of the three boundary edges.
State the absolute maximum and minimum.

Hint

For boundary edges, substitute the parametric form of each edge into

f

and find the critical points of the resulting single-variable function. Compare all values, interior critical point values and all boundary extrema,to find the absolute extrema.

View Solution

Solution to question 1 (interior critical points):

f_x = 4x - 2y - 4 = 0, \quad f_y = 2y - 2x = 0 \Rightarrow y = x

Substituting:

4x - 2x - 4 = 0 \Rightarrow x = 2, y = 2

. Point

(2,2)

is inside

D

(check:

2+2=4 > 3

, so

(2,2)

is outside the triangle). No interior critical point in

D

. Solution to question 2 (boundary): Edge 1: $y=0$ , $0\le x\le 3$ .

g(x)=2x^2-4x

g'(x)=4x-4=0\Rightarrow x=1

. Values:

g(0)=0

g(1)=-2

g(3)=6

. Edge 2: $x=0$ , $0\le y\le 3$ .

h(y)=y^2

. Values:

h(0)=0

h(3)=9

. Edge 3: $x+y=3$ , i.e. $y=3-x$ , $0\le x\le 3$ .

p(x) = 2x^2+(3-x)^2-2x(3-x)-4x = 2x^2+9-6x+x^2-6x+2x^2-4x = 5x^2-16x+9

p'(x)=10x-16=0\Rightarrow x=1.6

. Values:

p(0)=9

p(1.6)=5(2.56)-25.6+9=12.8-25.6+9=-3.8

p(3)=45-48+9=6

. Solution to question 3:

All candidate values: $0, -2, 6, 0, 9, 9, -3.8, 6$ .

\text{Absolute maximum} = 9,\text{ at }(0,3)

\text{Absolute minimum} = -3.8 = -\frac{19}{5},\text{ at }(1.6,\,1.4)