Derivative Definition - IB Mathematics Analysis and Approaches

The derivative of $f(x) = x^2 - 6x$ is defined by $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ . Find the value of $c$ such that the instantaneous rate of change of $f$ at $x=c$ is equal to 4.
Answer: $5$ The instantaneous rate of change of $f$ at $x=c$ is $f'(c)$ . We need to find the value of $c$ for which $f'(c) = 4$ . First, find the derivative of $f(x) = x^2 - 6x$ . Using the power rule, $f'(x) = 2x - 6$ . Now, set $f'(c) = 4$ : $2c - 6 = 4$ . Solving for $c$ , we get $2c = 10$ , which means $c = 5$ .
The expression $\lim_{h \to 0} \frac{\cos(\frac{\pi}{3}+h) - \frac{1}{2}}{h}$ represents the value of $f'(a)$ for some function $f$ and point $a$ . Find this value.
Answer: $-\frac{\sqrt{3}}{2}$ The given expression matches the definition of the derivative, $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ . By comparing the terms, we can identify the function as $f(x) = \cos(x)$ and the point as $a = \frac{\pi}{3}$ , since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ . The limit therefore represents the value of the derivative of $\cos(x)$ at $x = \frac{\pi}{3}$ . The derivative of $f(x) = \cos(x)$ is $f'(x) = -\sin(x)$ . Evaluating at $a = \frac{\pi}{3}$ , we get $f'(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$ .
The gradient of the tangent to the curve $y = \sqrt{x}$ at the point $(4, 2)$ is given by which limit?
Answer: $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$ An alternative definition for the derivative of a function $f(x)$ at a point $x=a$ is $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ . For the function $f(x) = \sqrt{x}$ at the point $(4, 2)$ , we have $a=4$ and $f(a) = f(4) = \sqrt{4} = 2$ . Substituting these values into the formula gives $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$ .
For the function $f(x) = x^3$ , let $R_{avg}$ be the average rate of change on the interval $[1, 3]$ and $R_{inst}$ be the instantaneous rate of change at $x=1$ , defined by $\lim_{h \to 0} \frac{(1+h)^3-1}{h}$ . Find the value of $R_{avg} - R_{inst}$ .
Answer: $10$ First, calculate the average rate of change: $R_{avg} = \frac{f(3) - f(1)}{3-1} = \frac{3^3 - 1^3}{2} = \frac{27 - 1}{2} = \frac{26}{2} = 13$ . Next, calculate the instantaneous rate of change at $x=1$ by finding the derivative $f'(x) = 3x^2$ and evaluating it at $x=1$ , which gives $R_{inst} = f'(1) = 3(1)^2 = 3$ . The question asks for $R_{avg} - R_{inst} = 13 - 3 = 10$ .
The number $e$ is defined such that the derivative of $f(x)=e^x$ at $x=0$ is 1. Which of the following limits expresses this definition?
Answer: $\lim_{h \to 0} \frac{e^{h} - 1}{h} = 1$ The derivative of $f(x)=e^x$ at $x=0$ is given by the limit $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$ . Substituting the function $f(x)=e^x$ , we get $f(0+h) = e^h$ and $f(0) = e^0 = 1$ . Therefore, the limit is $\lim_{h \to 0} \frac{e^h - 1}{h}$ . The statement that this derivative is equal to 1 is expressed as $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ .
The limit expression $\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}$ represents the derivative of a function $f(x)$ at a point $x=a$ . What are $f(x)$ and $a$ ?
Answer: $f(x) = x^2$ and $a=3$ The definition of the derivative of a function $f(x)$ at a point $x=a$ is $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ . Comparing this to the given limit, $\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}$ , we can identify $a=3$ . This means $f(a+h) = f(3+h) = (3+h)^2$ and $f(a) = f(3) = 9$ . From these, we can deduce that the function is $f(x) = x^2$ , since $f(3) = 3^2 = 9$ .
Given a function $f(x)$ such that $f'(a)$ exists, which of the following limits is also equal to $f'(a)$ ?
Answer: $\lim_{h \to 0} \frac{f(a) - f(a-h)}{h}$ The standard definition is $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ . Let's analyze the options. A: $\lim_{h \to 0} 2 \cdot \frac{f(a+2h) - f(a)}{2h} = 2f'(a)$ . B: $\frac{1}{2} \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \frac{1}{2}f'(a)$ . C: Let $k = -h$ . As $h \to 0$ , $k \to 0$ . The expression becomes $\lim_{k \to 0} \frac{f(a) - f(a+k)}{-k} = \lim_{k \to 0} \frac{-(f(a+k)-f(a))}{-k} = \lim_{k \to 0} \frac{f(a+k)-f(a)}{k} = f'(a)$ . This is correct. D: This limit equals $2f'(a)$ .
Find the value of the limit $\lim_{x \to 2} \frac{x^4 - 16}{x-2}$ .
Answer: $32$ This limit is in the form of the alternative definition of the derivative, $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$ . Here, $f(x) = x^4$ and $a=2$ , since $f(2) = 2^4 = 16$ . The limit is therefore equal to $f'(2)$ . The derivative is $f'(x) = 4x^3$ . Evaluating at $x=2$ gives $f'(2) = 4(2^3) = 4(8) = 32$ . Alternatively, one can factor the numerator as a difference of squares: $\frac{(x^2-4)(x^2+4)}{x-2} = \frac{(x-2)(x+2)(x^2+4)}{x-2}$ . Cancelling $(x-2)$ and substituting $x=2$ gives $(2+2)(2^2+4) = 4(8) = 32$ .
Find the value of the limit $\lim_{k \to 0} \frac{(a+3k)^2 - a^2}{k}$ .
Answer: $6a$ We can evaluate this limit algebraically. First, expand the numerator: $(a+3k)^2 - a^2 = (a^2 + 6ak + 9k^2) - a^2 = 6ak + 9k^2$ . Now, substitute this back into the limit expression: $\lim_{k \to 0} \frac{6ak + 9k^2}{k}$ . We can factor out $k$ from the numerator: $\lim_{k \to 0} \frac{k(6a + 9k)}{k}$ . For $k \neq 0$ , we can cancel $k$ : $\lim_{k \to 0} (6a + 9k)$ . Now, we can substitute $k=0$ to evaluate the limit, which gives $6a + 9(0) = 6a$ . Alternatively, this limit is a variation of the derivative definition for $f(x)=x^2$ at $x=a$ . It represents $3f'(a) = 3(2a) = 6a$ .
The quantity $\lim_{h \to 0} \frac{A(t+h) - A(t)}{h}$ , where $A(t)$ is the area of a circular oil slick in $m^2$ at time $t$ in seconds, represents:
Answer: the instantaneous rate of change of the area at time $t$ . The expression $\frac{A(t+h) - A(t)}{h}$ represents the average rate of change of the area $A$ over a time interval of length $h$ . The limit of this expression as $h \to 0$ is the definition of the derivative, $A'(t)$ , which represents the instantaneous rate of change of the area with respect to time at the specific moment $t$ .
If $y=f(x)$ is a differentiable function, what is the correct interpretation of the statement $f'(2) = 5$ ?
Answer: The tangent line to the graph of $y=f(x)$ at $x=2$ has a gradient of 5. The derivative of a function at a point, $f'(a)$ , has two main interpretations: the instantaneous rate of change of the function at $x=a$ , and the gradient (slope) of the tangent line to the graph of the function at the point $(a, f(a))$ . Therefore, $f'(2)=5$ means the gradient of the tangent line at $x=2$ is 5. Option A is $f(2)=5$ . Option C describes an average rate of change. Option D describes a linear function, which is not necessarily true for $f(x)$ in general.
Which of the following limits represents the instantaneous rate of change of the function $f(x) = \frac{1}{x}$ at $x=2$ ?
Answer: $\lim_{h \to 0} \frac{\frac{1}{2+h} - \frac{1}{2}}{h}$ The instantaneous rate of change of a function $f(x)$ at a point $x=a$ is given by the definition of the derivative, $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ . For the function $f(x) = \frac{1}{x}$ at $a=2$ , we substitute these into the formula to get $\lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0} \frac{\frac{1}{2+h} - \frac{1}{2}}{h}$ .
Let $g(x)$ be an even, differentiable function. Find the value of $\lim_{h \to 0} \frac{g(h) - g(-h)}{h}$ .
Answer: $0$ A function $g(x)$ is even if $g(x) = g(-x)$ for all $x$ in its domain. Therefore, for the numerator of the given expression, we have $g(h) - g(-h) = g(h) - g(h) = 0$ . The expression inside the limit is $\frac{0}{h}$ , which is 0 for all $h \neq 0$ . The limit of 0 as $h \to 0$ is 0.
Let $f(x) = \frac{1}{x+2}$ . The difference quotient $\frac{f(x+h) - f(x)}{h}$ can be simplified to which of the following expressions before taking the limit as $h \to 0$ ?
Answer: $\frac{-1}{(x+h+2)(x+2)}$ First, set up the difference quotient: $\frac{\frac{1}{x+h+2} - \frac{1}{x+2}}{h}$ . To simplify, find a common denominator for the fractions in the numerator: $\frac{\frac{(x+2) - (x+h+2)}{(x+h+2)(x+2)}}{h}$ . Simplify the numerator of the complex fraction: $\frac{\frac{x+2-x-h-2}{(x+h+2)(x+2)}}{h} = \frac{\frac{-h}{(x+h+2)(x+2)}}{h}$ . This is equivalent to $\frac{-h}{(x+h+2)(x+2)} \cdot \frac{1}{h}$ . Cancelling the $h$ terms gives the simplified expression $\frac{-1}{(x+h+2)(x+2)}$ .
The gradient of the tangent to the curve $y=f(x)$ at $x=1$ is given by the limit $\lim_{h \to 0} \frac{3(1+h)^2 - 3}{h}$ . Given that $f(1)=3$ , find the equation of the normal line to the curve at $x=1$ .
Answer: $y = -\frac{1}{6}x + \frac{19}{6}$ Step 1: Find the gradient of the tangent, $m_t$ , by evaluating the limit. $m_t = \lim_{h \to 0} \frac{3(1+2h+h^2) - 3}{h} = \lim_{h \to 0} \frac{3+6h+3h^2 - 3}{h} = \lim_{h \to 0} \frac{6h+3h^2}{h} = \lim_{h \to 0} (6+3h) = 6$ . Step 2: Find the gradient of the normal, $m_n$ . The normal is perpendicular to the tangent, so $m_n = -\frac{1}{m_t} = -\frac{1}{6}$ . Step 3: Find the equation of the normal line. The line passes through the point $(1, f(1)) = (1, 3)$ and has gradient $-\frac{1}{6}$ . Using the point-slope form $y - y_1 = m(x - x_1)$ : $y - 3 = -\frac{1}{6}(x - 1)$ . Rearranging to slope-intercept form: $y = -\frac{1}{6}x + \frac{1}{6} + 3 = -\frac{1}{6}x + \frac{1}{6} + \frac{18}{6} = -\frac{1}{6}x + \frac{19}{6}$ .
Find the gradient of the tangent to the curve $y=\sqrt{x}$ at $x=9$ by evaluating the limit $\lim_{h \to 0} \frac{\sqrt{9+h}-3}{h}$ .
Answer: $1/6$ To evaluate this limit, we multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{9+h}+3$ . This gives: $\lim_{h \to 0} \frac{\sqrt{9+h}-3}{h} \cdot \frac{\sqrt{9+h}+3}{\sqrt{9+h}+3} = \lim_{h \to 0} \frac{(9+h) - 9}{h(\sqrt{9+h}+3)}$ . Simplifying the numerator gives $\lim_{h \to 0} \frac{h}{h(\sqrt{9+h}+3)}$ . We can cancel $h$ (since $h \neq 0$ ), leaving $\lim_{h \to 0} \frac{1}{\sqrt{9+h}+3}$ . Now, we can substitute $h=0$ to find the limit: $\frac{1}{\sqrt{9+0}+3} = \frac{1}{3+3} = \frac{1}{6}$ .
Consider the function $f(x) = |x-3|$ . The derivative at a point $x=a$ is defined as $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ . For which value of $a$ does this limit fail to exist?
Answer: $3$ The function $f(x) = |x-3|$ has a sharp corner (a cusp) at $x=3$ . A function is not differentiable at a sharp corner. To verify this with the limit definition, we check the left-hand and right-hand limits at $a=3$ . The right-hand limit is $\lim_{h \to 0^+} \frac{|(3+h)-3| - 0}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = 1$ . The left-hand limit is $\lim_{h \to 0^-} \frac{|(3+h)-3| - 0}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$ . Since the left-hand and right-hand limits are not equal, the limit does not exist at $a=3$ .
The height $H$ in metres of a ball thrown vertically upwards is modelled by $H(t) = 20t - 5t^2$ , where $t$ is time in seconds. The instantaneous velocity of the ball at $t=1$ is defined by $\lim_{h \to 0} \frac{H(1+h) - H(1)}{h}$ . Calculate this velocity.
Answer: $10$ m/s We need to evaluate the limit for $H(t) = 20t - 5t^2$ at $t=1$ . First, find $H(1) = 20(1) - 5(1)^2 = 15$ . Next, find $H(1+h) = 20(1+h) - 5(1+h)^2 = 20 + 20h - 5(1 + 2h + h^2) = 20 + 20h - 5 - 10h - 5h^2 = 15 + 10h - 5h^2$ . Now, set up the difference quotient: $\frac{H(1+h) - H(1)}{h} = \frac{(15 + 10h - 5h^2) - 15}{h} = \frac{10h - 5h^2}{h}$ . Factor out $h$ : $\frac{h(10 - 5h)}{h} = 10 - 5h$ . Finally, take the limit: $\lim_{h \to 0} (10 - 5h) = 10$ . The instantaneous velocity at $t=1$ is 10 m/s.
The line $y=8x-5$ is tangent to the curve $y=kx^2$ at some point. The gradient of the tangent at a point $x$ on the curve is given by $\lim_{h \to 0} \frac{k(x+h)^2 - kx^2}{h}$ . Find the value of $k$ .
Answer: $16/5$ The gradient of the tangent line $y=8x-5$ is 8. The derivative of $y=kx^2$ , which is given by the limit, must be equal to 8 at the point of tangency. The derivative is $y' = 2kx$ . So, $2kx=8$ , which simplifies to $kx=4$ . At the point of tangency, the coordinates must be the same for the line and the curve, so $kx^2 = 8x-5$ . We can substitute $kx=4$ into this equation: $(kx)x = 8x-5 \implies 4x = 8x-5$ . Solving for $x$ gives $4x=5$ , so $x=5/4$ . Now we can find $k$ using $kx=4$ : $k(5/4)=4 \implies k = 16/5$ .

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