1NumericalGradient vectorAnswer

(question text pending review) # Solution

Part 1: Gradient and Directional Derivative [5 marks]

Definition of Gradient: The gradient of a scalar function $f(x,y,z)$ is defined as:

(question text pending review) # Solution

Part 1: Gradient and Directional Derivative [5 marks]

Definition of Gradient: The gradient of a scalar function $f(x,y,z)$ is defined as: $$

Part 1: - Function: $f(x,y,z) = x^3 - xy^2 + z$ - Point: $P(1,0,0)$ - Direction: $\vec{v} = 2\vec{i} - \vec{j} + \vec{k}$ Part 2: - Region bounded by $y = \sqrt{x}$ , line $y = 1$ , and $x = 4$ - Axis of revolution: $y = 1$ All data presen...

Part 1: - Function: $f(x,y,z) = x^3 - xy^2 + z$ - Point: $P(1,0,0)$ - Direction: $\vec{v} = 2\vec{i} - \vec{j} + \vec{k}$ Part 2: - Region bounded by $y = \sqrt{x}$, line $y = 1$, and $x = 4$ - Axis of revolution: $y = 1$ All data presen...

210 marksNumericalIntegration by partsAnswer

Evaluate $\int_{0}^{\frac{\pi}{4}} \frac{dx}{1-\sin x}$ , Evaluate $\int x^2 \sin x \, dx$ . Solve the differential equation $\frac{dy}{dx} - \frac{3y}{x} = x$ , $x > 0$ . [5+5]

Evaluate $\int_{0}^{\frac{\pi}{4}} \frac{dx}{1-\sin x}$, Evaluate $\int x^2 \sin x , dx$. Solve the differential equation $\frac{dy}{dx} - \frac{3y}{x} = x$, $x > 0$. [5+5]

Given data: Integrand $\frac{1}{1-\sin x}$ , limits $0$ to $\pi/4$ . Multiply numerator and denominator by the conjugate $(1+\sin x)$ : $$\int0^{\pi/4} \frac{1+\sin x}{(1-\sin x)(1+\sin x)},dx = \int0^{\pi/4} \frac{1+\sin x}{1-\sin^2 x},d...

Given data: Integrand $\frac{1}{1-\sin x}$, limits $0$ to $\pi/4$. Multiply numerator and denominator by the conjugate $(1+\sin x)$: $$\int0^{\pi/4} \frac{1+\sin x}{(1-\sin x)(1+\sin x)},dx = \int0^{\pi/4} \frac{1+\sin x}{1-\sin^2 x},d...

310 marksNumericalRolle's theoremAnswer

State Rolle's Theorem and show that $x^3 + 3x + 1 = 0$ has exactly one real solution. Find the area of the region enclosed by the parabola $y = 2 - x^2$ and the line y = -x. [5+5]

Function/equation: $x^3 + 3x + 1 = 0$ If a function $f$ is: 1. continuous on the closed interval $[a, b]$ , 2. differentiable on the open interval $(a, b)$ , and 3. $f(a) = f(b)$ , then there exists at least one point $c \in (a, b)$ such ...

Function/equation: $x^3 + 3x + 1 = 0$ If a function $f$ is: 1. continuous on the closed interval $[a, b]$, 2. differentiable on the open interval $(a, b)$, and 3. $f(a) = f(b)$, then there exists at least one point $c \in (a, b)$ such ...

45 marksAbsolute value functionsAnswer

Define absolute value function and Sketch the graph of absolute value. [5]

Absolute Value Function: Definition and Graph

Definition

The absolute value function is defined as:

$f(x) = |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$

In other words, the absolute value of a number is its distance from zero on the number line, always expressed as a non-negative value.

Key characteristics:

Domain: All real numbers (ℝ)
Range: [0, ∞) or all non-negative real numbers
The function is always non-negative: |x| ≥ 0 for all x ∈ ℝ

Graph of f(x) = |x|

        y
        |
        |     /
        |    /
        |   /
        |  /
    ____|_/______ x
       /|
      / |
     /  |
    /   |

Detailed sketch:

The graph consists of two linear pieces:

For x ≥ 0: the line y = x (slope = 1)
For x < 0: the line y = -x (slope = -1)

Important points:

Vertex at origin: (0, 0)
The graph is V-shaped
Symmetric about the y-axis (even function)
The two rays meet at a sharp point at the origin
Passes through points like (-2, 2), (-1, 1), (0, 0), (1, 1), (2, 2)

Properties:

Continuous everywhere
Not differentiable at x = 0 (sharp corner)
Decreasing on (-∞, 0] and increasing on [0, ∞)

Absolute Value Function: Definition and Graph

Definition

The absolute value function is defined as:

$$f(x) = |x| = \begin{cases} x & \text{if } x \geq 0 \ -x & \text{if } x < 0 \end{cases}$$

In other words, the absolute value of a number is its distance from zero on the number line, always expressed as a non-negative value.

Key characteristics:

Domain: All real numbers (ℝ)
Range: [0, ∞) or all non-negative real numbers
The function is always non-negative: |x| ≥ 0 for all x ∈ ℝ

Graph of f(x) = |x|

        y
        |
        |     /
        |    /
        |   /
        |  /
    ____|_/______ x
       /|
      / |
     /  |
    /   |

Detailed sketch:

The graph consists of two linear pieces:

For x ≥ 0: the line y = x (slope = 1)
For x < 0: the line y = -x (slope = -1)

Important points:

Vertex at origin: (0, 0)
The graph is V-shaped
Symmetric about the y-axis (even function)
The two rays meet at a sharp point at the origin
Passes through points like (-2, 2), (-1, 1), (0, 0), (1, 1), (2, 2)

Properties:

Continuous everywhere
Not differentiable at x = 0 (sharp corner)
Decreasing on (-∞, 0] and increasing on [0, ∞)

55 marksNumericalLimits at infinityAnswer

Find the limit of $\lim_{h\to\infty} \frac{\sqrt{6h+25-5}}{h^2}$ . [5]

Find the limit of $\lim_{h\to\infty} \frac{\sqrt{6h+25-5}}{h^2}$. [5]

The expression as literally rendered is garbled, but the most sensible interpretation is: $\lim{h \to \infty} \frac{\sqrt{6h + 25 - 5}}{h^2}$ - Numerator: $\sqrt{6h + 25 - 5} = \sqrt{6h + 20}$ - Denominator: $h^2$ - Limit variable:

The expression as literally rendered is garbled, but the most sensible interpretation is: $$\lim{h \to \infty} \frac{\sqrt{6h + 25 - 5}}{h^2}$$ - Numerator: $\sqrt{6h + 25 - 5} = \sqrt{6h + 20}$ - Denominator: $h^2$ - Limit variable:

65 marksNumericalIntegral testAnswer

State integral test and apply it to test the convergence of the series $\sum_{n=1}^{\infty}\frac{1}{n^2+1}$ . [5]

State integral test and apply it to test the convergence of the series $\sum_{n=1}^{\infty}\frac{1}{n^2+1}$. [5]

Series: $\displaystyle\sum{n=1}^{\infty}\frac{1}{n^2+1}$ - Associated function: $f(x)=\dfrac{1}{x^2+1}$ , tested on $[1,\infty)$ Let $f(x)$ be a function that is continuous, positive, and monotonically decreasing on $[1,\infty)$ , and su...

Series: $\displaystyle\sum{n=1}^{\infty}\frac{1}{n^2+1}$ - Associated function: $f(x)=\dfrac{1}{x^2+1}$, tested on $[1,\infty)$ Let $f(x)$ be a function that is continuous, positive, and monotonically decreasing on $[1,\infty)$, and su...

75 marksNumericalTaylor seriesAnswer

Find the Taylor's Series generated by $f(x) = \frac{1}{x}$ at $a = 2$ . Where, if anywhere, does the series converge to $\frac{1}{x}$ ? [5]

Find the Taylor's Series generated by $f(x) = \frac{1}{x}$ at $a = 2$. Where, if anywhere, does the series converge to $\frac{1}{x}$? [5]

Taylor Series for f(x) = 1/x at a = 2

Given data

Function: $f(x) = \dfrac{1}{x}$
Center: $a = 2$

Step 1: Derivatives of f(x) = 1/x

$f(x) = x^{-1}$ $f'(x) = -x^{-2}$ $f''(x) = 2x^{-3}$ $f'''(x) = -6x^{-4}$

General pattern: $f^{(n)}(x) = (-1)^n \, n! \, x^{-(n+1)}$

Step 2: Evaluate at a = 2

$f^{(n)}(2) = (-1)^n \, n! \, 2^{-(n+1)} = \frac{(-1)^n \, n!}{2^{n+1}}$

Values:

$f(2) = \dfrac{1}{2}$
$f'(2) = -\dfrac{1}{4}$
$f''(2) = \dfrac{2}{8} = \dfrac{1}{4}$ (coefficient term $\frac{f''(2)}{2!} = \frac{1}{8}$ )
$f'''(2) = -\dfrac{6}{16} = -\dfrac{3}{8}$

Step 3: Taylor series

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n = \sum_{n=0}^{\infty} \frac{(-1)^n \, n!}{n! \, 2^{n+1}}(x-2)^n$

$\boxed{\frac{1}{x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(x-2)^n = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3 + \cdots}$

Convergence Analysis

This is a geometric series. Write: $\frac{1}{x} = \frac{1}{2 + (x-2)} = \frac{1}{2}\cdot \frac{1}{1 + \frac{x-2}{2}} = \frac{1}{2}\sum_{n=0}^\infty \left(-\frac{x-2}{2}\right)^n$

which matches the series above.

Ratio Test: $\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}/2^{n+2}}{(x-2)^n/2^{n+1}}\right| = \frac{|x-2|}{2}$

Convergence requires: $\frac{|x-2|}{2} < 1 \implies |x-2| < 2 \implies 0 < x < 4$

Endpoint check: At $x=0$ and $x=4$ , $\left|\frac{x-2}{2}\right| = 1$ , so terms do not tend to 0; the series diverges at both endpoints.

Conclusion:

Because a geometric series converges to $\dfrac{1}{1-r}$ (here $\frac{1}{2}\cdot\frac{1}{1+(x-2)/2} = \frac{1}{x}$ ) whenever it converges, the sum equals $\dfrac{1}{x}$ throughout the interval of convergence:

$\text{The series converges to } \frac{1}{x} \text{ for } 0 < x < 4.$

Taylor Series for f(x) = 1/x at a = 2

Given data

Function: $f(x) = \dfrac{1}{x}$
Center: $a = 2$

Step 1: Derivatives of f(x) = 1/x

$$f(x) = x^{-1}$$ $$f'(x) = -x^{-2}$$ $$f''(x) = 2x^{-3}$$ $$f'''(x) = -6x^{-4}$$

General pattern: $$f^{(n)}(x) = (-1)^n , n! , x^{-(n+1)}$$

Step 2: Evaluate at a = 2

$$f^{(n)}(2) = (-1)^n , n! , 2^{-(n+1)} = \frac{(-1)^n , n!}{2^{n+1}}$$

Values:

$f(2) = \dfrac{1}{2}$
$f'(2) = -\dfrac{1}{4}$
$f''(2) = \dfrac{2}{8} = \dfrac{1}{4}$ (coefficient term $\frac{f''(2)}{2!} = \frac{1}{8}$)
$f'''(2) = -\dfrac{6}{16} = -\dfrac{3}{8}$

Step 3: Taylor series

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n = \sum_{n=0}^{\infty} \frac{(-1)^n , n!}{n! , 2^{n+1}}(x-2)^n$$

$$\boxed{\frac{1}{x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(x-2)^n = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3 + \cdots}$$

Convergence Analysis

This is a geometric series. Write: $$\frac{1}{x} = \frac{1}{2 + (x-2)} = \frac{1}{2}\cdot \frac{1}{1 + \frac{x-2}{2}} = \frac{1}{2}\sum_{n=0}^\infty \left(-\frac{x-2}{2}\right)^n$$

which matches the series above.

Ratio Test: $$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}/2^{n+2}}{(x-2)^n/2^{n+1}}\right| = \frac{|x-2|}{2}$$

Convergence requires: $$\frac{|x-2|}{2} < 1 \implies |x-2| < 2 \implies 0 < x < 4$$

Endpoint check: At $x=0$ and $x=4$, $\left|\frac{x-2}{2}\right| = 1$, so terms do not tend to 0; the series diverges at both endpoints.

Conclusion:

Because a geometric series converges to $\dfrac{1}{1-r}$ (here $\frac{1}{2}\cdot\frac{1}{1+(x-2)/2} = \frac{1}{x}$) whenever it converges, the sum equals $\dfrac{1}{x}$ throughout the interval of convergence:

$$\text{The series converges to } \frac{1}{x} \text{ for } 0 < x < 4.$$

85 marksNumericalImplicit differentiationAnswer

Define implicit differentiation and find the slope of the circle $x^2 + y^2 = 25$ at the point (3, -4). [5]

STEP 1 - EXTRACT

Given data:

Curve (implicit relation): $x^2 + y^2 = 25$ (a circle of radius 5 centered at origin)
Point of interest: $(3, -4)$

Check that the point lies on the circle: $3^2 + (-4)^2 = 9 + 16 = 25$ ✓

All required data present.

STEP 2 - SOLVE

Definition of Implicit Differentiation

Implicit differentiation is the process of finding the derivative $\dfrac{dy}{dx}$ when $y$ is defined implicitly as a function of $x$ through an equation $F(x, y) = 0$ , rather than explicitly as $y = f(x)$ .

Procedure:

Differentiate both sides of the equation with respect to $x$ .
Treat $y$ as a function of $x$ , applying the chain rule to every term containing $y$ (i.e., $\frac{d}{dx}[f(y)] = f'(y)\frac{dy}{dx}$ ).
Algebraically solve the resulting equation for $\dfrac{dy}{dx}$ .

Finding the Slope

Step 1: Differentiate both sides with respect to $x$

$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$

$2x + 2y\frac{dy}{dx} = 0$

Step 2: Solve for $\dfrac{dy}{dx}$

$2y\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$

Step 3: Substitute the point $(3, -4)$

$\frac{dy}{dx}\bigg|_{(3,-4)} = -\frac{3}{-4} = \frac{3}{4}$

Final Result

$\boxed{\text{Slope} = \frac{3}{4} = 0.75}$

STEP 1 - EXTRACT

Given data:

Curve (implicit relation): $x^2 + y^2 = 25$ (a circle of radius 5 centered at origin)
Point of interest: $(3, -4)$

Check that the point lies on the circle: $3^2 + (-4)^2 = 9 + 16 = 25$ ✓

All required data present.

STEP 2 - SOLVE

Definition of Implicit Differentiation

Implicit differentiation is the process of finding the derivative $\dfrac{dy}{dx}$ when $y$ is defined implicitly as a function of $x$ through an equation $F(x, y) = 0$, rather than explicitly as $y = f(x)$.

Procedure:

Differentiate both sides of the equation with respect to $x$.
Treat $y$ as a function of $x$, applying the chain rule to every term containing $y$ (i.e., $\frac{d}{dx}[f(y)] = f'(y)\frac{dy}{dx}$).
Algebraically solve the resulting equation for $\dfrac{dy}{dx}$.

Finding the Slope

Step 1: Differentiate both sides with respect to $x$

$$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$$

$$2x + 2y\frac{dy}{dx} = 0$$

Step 2: Solve for $\dfrac{dy}{dx}$

$$2y\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$$

Step 3: Substitute the point $(3, -4)$

$$\frac{dy}{dx}\bigg|_{(3,-4)} = -\frac{3}{-4} = \frac{3}{4}$$

Final Result

$$\boxed{\text{Slope} = \frac{3}{4} = 0.75}$$

95 marksNumericalPartial derivativesAnswer

Define partial derivative and find the value of $\frac{\partial f}{\partial x}$ & $\frac{\partial f}{\partial y}$ at the point (4, -5) if $f(x,y) = x^3 + 3xy + y - 1$ . [5]

Define partial derivative and find the value of $\frac{\partial f}{\partial x}$ & $\frac{\partial f}{\partial y}$ at the point (4, -5) if $f(x,y) = x^3 + 3xy + y - 1$. [5]

Model Answer: Partial Derivatives

STEP 1 - Given Data

Function: $f(x, y) = x^3 + 3xy + y - 1$
Point of evaluation: $(x, y) = (4, -5)$
Required: $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ at $(4, -5)$

Definition of Partial Derivative

A partial derivative of a function $f(x, y)$ with respect to one variable is its rate of change with respect to that variable while all other variables are held constant.

$\frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h}$

$\frac{\partial f}{\partial y} = \lim_{k \to 0} \frac{f(x, y+k) - f(x, y)}{k}$

STEP 2 - Solution

Finding $\dfrac{\partial f}{\partial x}$ (treat $y$ constant)

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\big(x^3 + 3xy + y - 1\big) = 3x^2 + 3y + 0 - 0 = 3x^2 + 3y$

Finding $\dfrac{\partial f}{\partial y}$ (treat $x$ constant)

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\big(x^3 + 3xy + y - 1\big) = 0 + 3x + 1 - 0 = 3x + 1$

Evaluating at $(4, -5)$

$\frac{\partial f}{\partial x}\bigg|_{(4,-5)} = 3(4)^2 + 3(-5) = 48 - 15 = 33$

$\frac{\partial f}{\partial y}\bigg|_{(4,-5)} = 3(4) + 1 = 12 + 1 = 13$

Final Answer

$\boxed{\dfrac{\partial f}{\partial x}\bigg|_{(4,-5)} = 33, \qquad \dfrac{\partial f}{\partial y}\bigg|_{(4,-5)} = 13}$

Model Answer: Partial Derivatives

STEP 1 - Given Data

Function: $f(x, y) = x^3 + 3xy + y - 1$
Point of evaluation: $(x, y) = (4, -5)$
Required: $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ at $(4, -5)$

Definition of Partial Derivative

A partial derivative of a function $f(x, y)$ with respect to one variable is its rate of change with respect to that variable while all other variables are held constant.

$$\frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h}$$

$$\frac{\partial f}{\partial y} = \lim_{k \to 0} \frac{f(x, y+k) - f(x, y)}{k}$$

STEP 2 - Solution

Finding $\dfrac{\partial f}{\partial x}$ (treat $y$ constant)

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\big(x^3 + 3xy + y - 1\big) = 3x^2 + 3y + 0 - 0 = 3x^2 + 3y$$

Finding $\dfrac{\partial f}{\partial y}$ (treat $x$ constant)

$$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\big(x^3 + 3xy + y - 1\big) = 0 + 3x + 1 - 0 = 3x + 1$$

Evaluating at $(4, -5)$

$$\frac{\partial f}{\partial x}\bigg|_{(4,-5)} = 3(4)^2 + 3(-5) = 48 - 15 = 33$$

$$\frac{\partial f}{\partial y}\bigg|_{(4,-5)} = 3(4) + 1 = 12 + 1 = 13$$

Final Answer

$$\boxed{\dfrac{\partial f}{\partial x}\bigg|{(4,-5)} = 33, \qquad \dfrac{\partial f}{\partial y}\bigg|{(4,-5)} = 13}$$

105 marksNumericalTrigonometric integralsAnswer

Evaluate $\int_0^{\frac{\pi}{2}}(\sin 2x\cos 3x + \cos 2x\sin 3x)dx$ and $\int_{0}^{\frac{\pi}{4}} \frac{dx}{1 - \sin x}$ . [5]

Evaluate $\int_0^{\frac{\pi}{2}}(\sin 2x\cos 3x + \cos 2x\sin 3x)dx$ and $\int_{0}^{\frac{\pi}{4}} \frac{dx}{1 - \sin x}$. [5]

Integral 1: $\displaystyle\int0^{\pi/2}(\sin 2x\cos 3x + \cos 2x\sin 3x)\,dx$ - Integral 2: $\displaystyle\int0^{\pi/4}\frac{dx}{1-\sin x}$ --- Step 1: Apply sine addition formula $\sin A\cos B + \cos A\sin B = \sin(A+B)$ With

Integral 1: $\displaystyle\int0^{\pi/2}(\sin 2x\cos 3x + \cos 2x\sin 3x),dx$ - Integral 2: $\displaystyle\int0^{\pi/4}\frac{dx}{1-\sin x}$ --- Step 1: Apply sine addition formula $$\sin A\cos B + \cos A\sin B = \sin(A+B)$$ With

115 marksNumericalConcavity and inflection pointsAnswer

Determine the concavity of y = 3 + sin x on [0, 2π\piπ]. [5]

Function: $y = 3 + \sin x$ - Interval: $[0, 2\pi]$ $y' = \cos x$ $y'' = -\sin x$ Concave up requires $y'' 0$ : $-\sin x 0 \implies \sin x < 0$ On $[0, 2\pi]$ , $\sin x < 0$ for $x \in (\pi, 2\pi)$ . Concave down requires $y'' < 0$ : ...

Function: $y = 3 + \sin x$ - Interval: $[0, 2\pi]$ $$y' = \cos x$$ $$y'' = -\sin x$$ Concave up requires $y'' 0$: $$-\sin x 0 \implies \sin x < 0$$ On $[0, 2\pi]$, $\sin x < 0$ for $x \in (\pi, 2\pi)$. Concave down requires $y'' < 0$: ...

125 marksNumericalRoot testAnswer

Test for convergence of the series $\sum_{n=1}^{\infty}\left(\frac{1}{n+1}\right)^n$ . [5]

Test for convergence of the series $\sum_{n=1}^{\infty}\left(\frac{1}{n+1}\right)^n$. [5]

Series: $\displaystyle\sum{n=1}^{\infty}\left(\frac{1}{n+1}\right)^n$ - General term: $an = \left(\dfrac{1}{n+1}\right)^n$ - All terms positive, so ordinary convergence = absolute convergence. No data missing. For a positive-term serie...

Series: $\displaystyle\sum{n=1}^{\infty}\left(\frac{1}{n+1}\right)^n$ - General term: $an = \left(\dfrac{1}{n+1}\right)^n$ - All terms positive, so ordinary convergence = absolute convergence. No data missing. For a positive-term serie...

Basic Mathematics 2080 question paper

Part 1: Gradient and Directional Derivative [5 marks]

Part 1: Gradient and Directional Derivative [5 marks]

Absolute Value Function: Definition and Graph

Definition

Graph of f(x) = |x|

Absolute Value Function: Definition and Graph

Definition

Graph of f(x) = |x|

Taylor Series for f(x) = 1/x at a = 2

Given data

Step 1: Derivatives of f(x) = 1/x

Step 2: Evaluate at a = 2

Step 3: Taylor series

Convergence Analysis

Taylor Series for f(x) = 1/x at a = 2

Given data

Step 1: Derivatives of f(x) = 1/x

Step 2: Evaluate at a = 2

Step 3: Taylor series

Convergence Analysis

STEP 1 - EXTRACT

STEP 2 - SOLVE

Definition of Implicit Differentiation

Finding the Slope

Final Result

STEP 1 - EXTRACT

STEP 2 - SOLVE

Definition of Implicit Differentiation

Finding the Slope

Final Result

Model Answer: Partial Derivatives

STEP 1 - Given Data

Definition of Partial Derivative

STEP 2 - Solution

Finding ∂f∂x\dfrac{\partial f}{\partial x}∂x∂f​ (treat yyy constant)

Finding ∂f∂y\dfrac{\partial f}{\partial y}∂y∂f​ (treat xxx constant)

Evaluating at (4,−5)(4, -5)(4,−5)

Final Answer

Model Answer: Partial Derivatives

STEP 1 - Given Data

Definition of Partial Derivative

STEP 2 - Solution

Finding $\dfrac{\partial f}{\partial x}$ (treat $y$ constant)

Finding $\dfrac{\partial f}{\partial y}$ (treat $x$ constant)

Evaluating at $(4, -5)$

Final Answer

Part 1: Gradient and Directional Derivative [5 marks]

Absolute Value Function: Definition and Graph

Definition

Graph of f(x) = |x|

Taylor Series for f(x) = 1/x at a = 2

Given data

Step 1: Derivatives of f(x) = 1/x

Step 2: Evaluate at a = 2

Step 3: Taylor series

Convergence Analysis

STEP 1 - EXTRACT

STEP 2 - SOLVE

Definition of Implicit Differentiation

Finding the Slope

Final Result

Model Answer: Partial Derivatives

STEP 1 - Given Data

Definition of Partial Derivative

STEP 2 - Solution

Finding ∂f∂x\dfrac{\partial f}{\partial x}∂x∂f​ (treat yyy constant)

Finding ∂f∂y\dfrac{\partial f}{\partial y}∂y∂f​ (treat xxx constant)

Evaluating at (4,−5)(4, -5)(4,−5)

Final Answer

Finding $\dfrac{\partial f}{\partial x}$ (treat $y$ constant)

Finding $\dfrac{\partial f}{\partial y}$ (treat $x$ constant)

Evaluating at $(4, -5)$

Finding $\dfrac{\partial f}{\partial x}$ (treat $y$ constant)

Finding $\dfrac{\partial f}{\partial y}$ (treat $x$ constant)

Evaluating at $(4, -5)$