Derivative Definition & Notation - Calculus 1

1. Let $f(x)$ be a differentiable function. The notation $\frac{d}{dt}[f(x)]$ represents:
   Answer: The derivative of $f(x)$ with respect to $t$, which is 0 if $x$ and $t$ are independent. The Leibniz notation \frac{d}{dt}[...]\] asks for the derivative of the expression inside the brackets with respect to the variable \(t. If the variable $x$ does not depend on $t$, then $f(x)$ is treated as a constant with respect to $t$, and its derivative is 0. If $x$ were a function of $t$, the chain rule would be required.
2. Consider the piecewise function $f(x) = \begin{cases} x^2 & \text{if } x \le 1 \\ 2x-1 & \text{if } x > 1 \end{cases}$. To determine if $f'(1)$ exists, one must evaluate the left and right limits of $\frac{f(1+h)-f(1)}{h}$ as $h \to 0$. What can be concluded?
   Answer: The left and right limits of the difference quotient are both equal to 2. First, note $f(1)=1^2=1$. The left-hand derivative is $\lim_{h \to 0^-} \frac{(1+h)^2 - 1}{h} = \lim_{h \to 0^-} \frac{1+2h+h^2-1}{h} = \lim_{h \to 0^-} (2+h) = 2$. The right-hand derivative is $\lim_{h \to 0^+} \frac{(2(1+h)-1) - 1}{h} = \lim_{h \to 0^+} \frac{2+2h-2}{h} = \lim_{h \to 0^+} 2 = 2$. Since the left-hand and right-hand limits of the difference quotient are equal, the derivative $f'(1)$ exists and equals 2.
3. Which of the following limits gives the slope of the tangent line to the graph of $f(x) = \ln(x)$ at the point $(e, 1)$?
   Answer: $\lim_{h \to 0} \frac{\ln(e+h) - 1}{h}$ The slope of the tangent line at a point $(a, f(a))$ is given by the derivative $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$. Here, the function is $f(x) = \ln(x)$ and the point is $(e, 1)$, so $a=e$ and $f(a)=1$. Substituting these values into the definition gives $\lim_{h \to 0} \frac{\ln(e+h) - 1}{h}$.
4. Given that $\lim_{x \to c} \frac{g(x)-g(c)}{x-c} = L$, where $L$ is a finite number. What must be the value of $\lim_{x \to c} g(x)$?
   Answer: $g(c)$ The given information $\lim_{x \to c} \frac{g(x)-g(c)}{x-c} = L$ is the definition of the derivative $g'(c) = L$. For a function to be differentiable at a point $c$, it must be continuous at $c$. The definition of continuity at $c$ is that $\lim_{x \to c} g(x) = g(c)$.
5. Let $g(x) = x^3$. Which of the following limits represents $g'(2)$?
   Answer: $\lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$ The derivative of a function $g(x)$ at a point $a$ is defined as $g'(a) = \lim_{h \to 0} \frac{g(a+h) - g(a)}{h}$. For $g(x)=x^3$ and $a=2$, we have $g(2) = 2^3 = 8$. Substituting into the definition gives $g'(2) = \lim_{h \to 0} \frac{g(2+h) - g(2)}{h} = \lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$.
6. The existence of $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ implies the truth of which other statement about the function $f$ at $x=a$?
   Answer: $\lim_{x \to a} f(x) = f(a)$ The existence of the given limit means that the function $f$ is differentiable at $x=a$. A fundamental theorem in calculus states that if a function is differentiable at a point, it must be continuous at that point. The definition of continuity at $x=a$ is that $\lim_{x \to a} f(x) = f(a)$.
7. The limit $L = \lim_{x \to \pi/2} \frac{\sin(x) - 1}{x - \pi/2}$ represents the slope of the tangent line to a curve $y=f(x)$ at a point $(a, f(a))$. What is the equation of this tangent line?
   Answer: $y = 1$ The limit is the definition of $f'(a)$ where $f(x)=\sin(x)$ and $a=\pi/2$. The value of the limit is the slope of the tangent line, $m = f'(\pi/2)$. Since $f'(x) = \cos(x)$, the slope is $m = \cos(\pi/2) = 0$. The point of tangency is $(a, f(a))$ which is $(\pi/2, \sin(\pi/2)) = (\pi/2, 1)$. Using the point-slope formula, $y - 1 = 0(x - \pi/2)$, which simplifies to $y = 1$.
8. The expression $\lim_{h \to 0} \frac{e^{2(x+h)} - e^{2x}}{h}$ represents the derivative of which function?
   Answer: $f(x) = e^{2x}$ The definition of the derivative is $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. By comparing the given expression with this definition, we can identify $f(x) = e^{2x}$. When we replace $x$ with $x+h$ in this function, we get $f(x+h) = e^{2(x+h)}$, which matches the first term in the numerator.
9. Suppose $f(1)=3$ and $f'(1)=2$. What is the value of $\lim_{x \to 1} \frac{f(x) - 3}{x-1}$?
   Answer: 2 The expression $\lim_{x \to 1} \frac{f(x) - 3}{x-1}$ matches the alternate definition of the derivative of $f$ at $x=1$. The definition is $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$. Here, $a=1$ and $f(a)=f(1)=3$. Therefore, the value of the limit is exactly $f'(1)$, which is given as 2.
10. If $y = f(x)$, the notation $\frac{dy}{dx}$ is defined by which limit?
    Answer: $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ When you encounter questions about derivative notation, you're being tested on understanding the fundamental definition of a derivative as the limit of difference quotients. The derivative $\frac{dy}{dx}$ represents the instantaneous rate of change of $y$ with respect to $x$. This is defined as the limit of the average rate of change as the change in $x$ approaches zero. If $y = f(x)$, then when $x$ changes by $\Delta x,y$ changes by $\Delta y = f(x + \Delta x) - f(x)$. The average rate of change is $\frac{\Delta y}{\Delta x}$, and taking the limit as $\Delta x \to 0$ gives us the instantaneous rate of change: $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$. This makes answer D correct. Answer A is incorrect because it compares $f(x)$ to $f(0)$ specifically, which would only give the average rate of change from $x = 0$ to some point $x$, not the instantaneous rate at a general point. Answer B has the limit variable wrong—we need $\Delta x \to 0$, not $\Delta y \to 0$. As $\Delta x$ shrinks, $\Delta y$ also approaches zero for continuous functions, but the limit must be taken with respect to the independent variable's change. Answer C represents $\lim_{x \to 0} \frac{f(x)}{x}$, which could be related to the derivative at $x = 0$ only if $f(0) = 0$, but this isn't the general definition. Remember: derivatives are always limits of difference quotients where the change in the independent variable approaches zero.
11. The limit $\lim_{h \to 0} \frac{(1+2h)^3 - 1}{h}$ represents the derivative of a function $g(x)$ at a point $x=a$. What is the value of this limit?
    Answer: $6$ The limit $\lim_{h \to 0} \frac{(1+2h)^3 - 1}{h}$ can be related to the derivative of $f(x)=x^3$ at $x=1$, which is $f'(1) = \lim_{k \to 0} \frac{(1+k)^3 - 1}{k}$. Let $k=2h$. Then $h=k/2$. As $h \to 0$, $k \to 0$. Substituting this into the given limit gives $\lim_{k \to 0} \frac{(1+k)^3 - 1}{k/2} = 2 \lim_{k \to 0} \frac{(1+k)^3 - 1}{k}$. This is $2$ times the derivative of $f(x)=x^3$ at $x=1$. Since $f'(x)=3x^2$, $f'(1)=3(1)^2=3$. Therefore, the value of the limit is $2 \times 3 = 6$. Alternatively, one can identify the limit as the derivative of $g(x)=(1+2x)^3$ at $x=0$. By the chain rule, $g'(x) = 3(1+2x)^2 \cdot 2 = 6(1+2x)^2$. Then $g'(0) = 6(1+0)^2 = 6$.
12. Let $f$ be a function differentiable at $x=a$. The expression $\lim_{h \to 0} \frac{f(a+h) - f(a-h)}{2h}$ is known as the symmetric difference quotient. If $f'(a)$ exists, what is the value of this limit?
    Answer: $f'(a)$ We can rewrite the expression by adding and subtracting $f(a)$ in the numerator: $\lim_{h \to 0} \frac{f(a+h) - f(a) - f(a-h) + f(a)}{2h}$ $= \lim_{h \to 0} \left[ \frac{f(a+h) - f(a)}{2h} + \frac{f(a) - f(a-h)}{2h} \right]$. This can be split into two limits: $\frac{1}{2} \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} + \frac{1}{2} \lim_{h \to 0} \frac{f(a-h) - f(a)}{-h}$. Both of these limits are equal to $f'(a)$. Therefore, the expression simplifies to $\frac{1}{2}f'(a) + \frac{1}{2}f'(a) = f'(a)$.
13. Let $f$ be a differentiable function with $f'(3) = -4$. What is the value of $\lim_{h \to 0} \frac{f(3+2h) - f(3)}{h}$?
    Answer: -8 This limit is a variation of the derivative definition $f'(a) = \lim_{k \to 0} \frac{f(a+k)-f(a)}{k}$. Let $k = 2h$. As $h \to 0$, $k \to 0$. We can rewrite the expression by substituting $h = k/2$: $\lim_{k \to 0} \frac{f(3+k) - f(3)}{k/2} = \lim_{k \to 0} 2 \cdot \frac{f(3+k) - f(3)}{k} = 2 \cdot f'(3)$. Since $f'(3) = -4$, the value of the limit is $2 \cdot (-4) = -8$.
14. Let $f$ be a function such that $f'(x)$ exists for all $x$. If $\lim_{x \to a} \frac{f(x)-f(a)}{x-a} = 0$, what can be concluded about the tangent line to the graph of $f$ at $x=a$?
    Answer: The tangent line is horizontal. The given limit is the definition of the derivative of $f$ at $x=a$. Therefore, $f'(a) = 0$. The value of the derivative at a point gives the slope of the tangent line at that point. A slope of 0 corresponds to a horizontal line.
15. Let $c$ be a constant. Which of the following expressions evaluates to 0, regardless of the function $g(x)$ (assuming $g(c)$ is defined)?
    Answer: $\frac{d}{dx}[g(c)]$ The notation $\frac{d}{dx}[g(c)]$ asks for the derivative with respect to $x$ of the value $g(c)$. Since $c$ is a constant, $g(c)$ is also a constant value. The derivative of any constant with respect to $x$ is 0. The other options represent $g'(c)$, which is not generally 0.
16. The line tangent to the graph of a differentiable function $f(x)$ at $x=-1$ is given by the equation $y=4x+3$. Which of the following statements must be true? I. $f(-1) = -1$ II. $\lim_{h \to 0} \frac{f(-1+h) + 1}{h} = 4$ III. $\lim_{x \to -1} \frac{f(x)+1}{x+1} = 4$
    Answer: I, II, and III The equation of the tangent line to $f(x)$ at $x=a$ is $y - f(a) = f'(a)(x-a)$. At $x=-1$, the tangent line must pass through the point $(-1, f(-1))$. Plugging $x=-1$ into the line equation $y=4x+3$ gives $y=4(-1)+3 = -1$. So, $f(-1)=-1$. Statement I is true. The slope of the tangent line is the derivative at the point of tangency. The slope of $y=4x+3$ is 4, so $f'(-1)=4$. Statement II is $\lim_{h \to 0} \frac{f(-1+h) + 1}{h}$. Since $f(-1)=-1$, this is equivalent to $\lim_{h \to 0} \frac{f(-1+h) - f(-1)}{h}$, which is the definition of $f'(-1)$. Since $f'(-1)=4$, statement II is true. Statement III is $\lim_{x \to -1} \frac{f(x)+1}{x+1}$. Since $f(-1)=-1$, this is $\lim_{x \to -1} \frac{f(x)-f(-1)}{x-(-1)}$, which is the alternate definition of $f'(-1)$. Since $f'(-1)=4$, statement III is true. Therefore, all three statements must be true.
17. The limit $L = \lim_{x \to 1} \frac{x^5 - 1}{x-1}$ represents $f'(a)$ for some function $f$ and some number $a$. What is the value of $L$?
    Answer: 5 This limit is in the form of the alternate definition of the derivative, $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$. By comparison, we can see that $f(x) = x^5$ and $a=1$. Note that $f(1) = 1^5 = 1$, which matches the numerator. The limit $L$ is therefore equal to $f'(1)$. First, find the derivative: $f'(x) = 5x^4$. Then evaluate at $x=1$: $f'(1) = 5(1)^4 = 5$.
18. If $f$ is a differentiable function, which of the following expressions is always equivalent to $f'(c)$?
    Answer: $\lim_{h \to 0} \frac{f(c+h) - f(c-h)}{2h}$ This expression is the symmetric difference quotient. We can rewrite it as $\lim_{h \to 0} \frac{f(c+h) - f(c) + f(c) - f(c-h)}{2h} = \lim_{h \to 0} \frac{1}{2} \left( \frac{f(c+h)-f(c)}{h} + \frac{f(c)-f(c-h)}{h} \right)$. The first part limits to $f'(c)$. For the second part, let $k=-h$. As $h \to 0$, $k \to 0$. It becomes $\lim_{k \to 0} \frac{f(c)-f(c+k)}{-k} = \lim_{k \to 0} \frac{f(c+k)-f(c)}{k} = f'(c)$. So the limit is $\frac{1}{2}(f'(c) + f'(c)) = f'(c)$. B is $-f'(c)$. C is $-f'(c)$. D is $2f'(c)$ by a change of variables $k=2h$.
19. The expression $\lim_{h \to 0} \frac{\cos(\frac{\pi}{3} + h) - \frac{1}{2}}{h}$ represents the derivative of a function $f(x)$ at a point $c$. What is the value of this limit?
    Answer: $-\frac{\sqrt{3}}{2}$ The given limit matches the definition of the derivative, $f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$. By comparing the expressions, we can identify the function $f(x) = \cos(x)$ and the point $c = \frac{\pi}{3}$, since $f(\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$. The value of the limit is $f'(\frac{\pi}{3})$. The derivative of $f(x)$ is $f'(x) = -\sin(x)$. Evaluating at $c = \frac{\pi}{3}$ gives $f'(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$.
20. If $f(x) = 2x^2$, which of the following is equivalent to the expression for $f'(x)$ derived from the definition?
    Answer: $\lim_{h \to 0} \frac{2(x+h)^2 - 2x^2}{h}$ The definition of the derivative of a function $f(x)$ is $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. For $f(x) = 2x^2$, we substitute $f(x+h) = 2(x+h)^2$ and $f(x) = 2x^2$ into the definition. This yields $\lim_{h \to 0} \frac{2(x+h)^2 - 2x^2}{h}$.

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