Mathematics I · Unit 3 · 4 hrs
Derivatives
Exam-focused notes for Derivatives (Mathematics I, MTH117): what the TU syllabus asks and how it has actually been tested, with 10 solved past questions from this unit.
What this unit covers
- Tangents and velocity
- Rate of change
- Review of derivative
- Differentiability of a function
- Mean value theorem
- Indeterminate forms and L'Hospital rule
Differentiability of a function
Where the function $f(x) = |x|$ is differentiable? Discuss.
A farmer has 1200 m. of fencing and wants to fence off a rectangular field that borders a straight river. He needs to fence along the river. What are the dimensions of the field that has the largest area? [5+5]
(a) Differentiability of f(x) = x - Function: $f(x) = x$ $f$ is differentiable at $a$ if the limit below exists: $$f'(a) = \lim{h \to 0} \frac{f(a+h) - f(a)}{h}$$ $$f(x) = x = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}$$ $f(x) = x \Rightarrow f'(x...
Full solved answer →Review of derivative
Find y' if $x^3 + y^3 = 6xy$. [5]
Equation (folium of Descartes): $x^3 + y^3 = 6xy$ Find: $y' = \dfrac{dy}{dx}$ Differentiate both sides with respect to $x$: $$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(6xy)$$ Left side (chain rule on $y^3$): $$3x^2 + 3y^2\,y'$$ Right side (produc...
Full solved answer →Find the derivative of $f(x) = \sqrt{x}$. State the domain of $f$. Estimate the area between the curve and the line $x = 0$ and $x = 2$ where curve is $y^2 = x$. [3+2+5]
- Function: $f(x) = \sqrt{x}$ - Curve: $y^2 = x$ - Bounds: $x = 0$ and $x = 2$ - Marks split: [3 + 2 + 5] --- (a) Derivative of $f(x) = \sqrt{x}$ Using first principles: $$f'(x) = \lim{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$ Multiply by the conjugate: $$...
Full solved answer →Tangents and velocity
Find the equation of the tangent at (1,3) to the curve $y = x^2 + 1$. [5]
- Curve: $y = x^2 + 1$ - Point of tangency: $(1, 3)$ Note on the point: Substituting $x = 1$ into the curve gives $y = 1^2 + 1 = 2$, so the point $(1, 3)$ does not actually lie on the curve $y = x^2 + 1$ (the curve passes through $(1, 2)$). This appears to ...
Full solved answer →Find the equation of tangent at (1,2) to the curve $y = 2x^3$. [5]
- Curve: $y = 2x^3$ - Point of tangency: $(x1, y1) = (1, 2)$ --- Substitute $x = 1$: $$y = 2(1)^3 = 2 \checkmark$$ The point $(1, 2)$ lies on the curve. --- $$\frac{dy}{dx} = 6x^2$$ --- $$m = 6(1)^2 = 6$$ --- Using $y - y1 = m(x - x1)$: $$y - 2 = 6(x - 1)$$...
Full solved answer →Mean value theorem
State Rolle's theorem and verify the theorem for $f(x) = x^2 - 9$, $x \in [-3,3]$. [5]
- Function: $f(x) = x^2 - 9$ - Interval: $[a, b] = [-3, 3]$ --- If a function $f$ satisfies: 1. $f$ is continuous on the closed interval $[a, b]$, 2. $f$ is differentiable on the open interval $(a, b)$, 3. $f(a) = f(b)$, then there exists at least one point...
Full solved answer →State Rolle's theorem and verify the Rolle's theorem for $f(x) = x^2 - 3x + 2$ in $[0, 3]$. [5]
- Function: $f(x) = x^2 - 3x + 2$ - Interval: $[0, 3]$, so $a = 0$, $b = 3$ If a function $f$ satisfies: 1. $f$ is continuous on the closed interval $[a, b]$, 2. $f$ is differentiable on the open interval $(a, b)$, 3. $f(a) = f(b)$, then there exists at lea...
Full solved answer →State Rolle's theorem and verify the Rolle's theorem for $f(x) = x^3 - x^2 - 6x + 2$ in [0, 3]. [5]
- Function: $f(x) = x^3 - x^2 - 6x + 2$ - Interval: $[0, 3]$, so $a = 0$, $b = 3$ If a function $f$ satisfies: 1. $f$ is continuous on the closed interval $[a, b]$ 2. $f$ is differentiable on the open interval $(a, b)$ 3. $f(a) = f(b)$ then there exists at ...
Full solved answer →Verify Mean value theorem of $f(x) = x^3 - 3x + 2$ for [-1,2]. [5]
- Function: $f(x) = x^3 - 3x + 2$ - Interval: $[a, b] = [-1, 2]$, so $a = -1$, $b = 2$ If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c \in (a,b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ $f(x) = x^3 -...
Full solved answer →Verify Mean value theorem of $f(x) = x^3 - 3x + 3$ for [-1,2]. [5]
- Function: $f(x) = x^3 - 3x + 3$ - Interval: $[a, b] = [-1, 2]$, so $a = -1$, $b = 2$ If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ $f(x)$ is a polynomial, he...
Full solved answer →Make Unit 3 stick
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