Determine the first derivative of each of the following functions. Print your answer in the box provided.
- \(\displaystyle f(x) = x \cos(5x) + x^2 + 8\).
- \(\displaystyle g(t) = e^{3t^2} \cdot \ln(8t + 1) + 2001\).
- \(\displaystyle F(y) = \frac{y}{y^2 + 1}\).
- \(\displaystyle H(u) = \arccos \left( \frac{u}{2} \right)\).
- \(\displaystyle h(x) = (5x^2 + 1)^{4x}\).
Solution:
- Differentiate \(x\cos(5x)\) with the product rule, and differentiate \(x^2+8\) term by term. Since \[ \frac{d}{dx}\bigl(x\cos(5x)\bigr)=1\cdot\cos(5x)+x\cdot(-5\sin(5x)), \] we obtain \[ \boxed{f'(x)=\cos(5x)-5x\sin(5x)+2x}. \]
- This is a product of \(e^{3t^2}\) and \(\ln(8t+1)\). First compute the derivatives of the factors: \[ \frac{d}{dt}e^{3t^2}=6te^{3t^2},\qquad \frac{d}{dt}\ln(8t+1)=\frac{8}{8t+1}. \] Now apply the product rule: \[ g'(t)=6te^{3t^2}\ln(8t+1)+e^{3t^2}\frac{8}{8t+1}. \] Factoring out \(e^{3t^2}\) gives \[ \boxed{g'(t)=e^{3t^2}\left(6t\ln(8t+1)+\frac{8}{8t+1}\right)}. \]
- Use the quotient rule with numerator \(N(y)=y\) and denominator \(D(y)=y^2+1\). Then \(N'(y)=1\) and \(D'(y)=2y\), so \[ F'(y)=\frac{N'D-ND'}{D^2} =\frac{(y^2+1)-y(2y)}{(y^2+1)^2} =\frac{1-y^2}{(y^2+1)^2}. \] Therefore \[ \boxed{F'(y)=\frac{1-y^2}{(y^2+1)^2}}. \]
- For \(H(u)=\arccos(u/2)\), use the chain rule. The derivative of \(\arccos(v)\) is \(-1/\sqrt{1-v^2}\), and the derivative of \(u/2\) is \(1/2\). Hence \[ H'(u)=-\frac{1/2}{\sqrt{1-u^2/4}}=-\frac{1}{\sqrt{4-u^2}}. \] So \[ \boxed{H'(u)=-\frac{1}{\sqrt{4-u^2}}}. \]
- Because the variable is in both the base and the exponent, use logarithmic differentiation. Let \[ y=(5x^2+1)^{4x}. \] Then \[ \ln y=4x\ln(5x^2+1). \] Differentiate both sides: \[ \frac{y'}{y}=4\ln(5x^2+1)+4x\cdot\frac{10x}{5x^2+1} =4\ln(5x^2+1)+\frac{40x^2}{5x^2+1}. \] Multiply by \(y\) to solve for \(y'\): \[ \boxed{h'(x)=(5x^2+1)^{4x}\left(4\ln(5x^2+1)+\frac{40x^2}{5x^2+1}\right)}. \]
Determine the most general anti-derivative for each of the following expressions. Print your answer in the box provided.
- \(\displaystyle \int (10x^4 - e^x + 2001)\ dx\).
Anti-Derivative:
- \(\displaystyle \int \left( \cos(8x) - \frac{1}{x} \right)\ dx\).
Anti-Derivative:
- \(\displaystyle \int x \sin(4x^2 - 1)\ dx\).
Anti-Derivative:
- \(\displaystyle \int \frac{x}{\sqrt{1+x^2}} dx\).
- \(\displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx\).
Solution:
- \(\displaystyle \int(10x^4-e^x+2001)dx=\boxed{2x^5-e^x+2001x+C}\).
- \(\displaystyle \int\left(\cos(8x)-\frac1x\right)dx=\boxed{\frac18\sin(8x)-\ln|x|+C}\).
- Let \(u=4x^2-1\), so \(du=8x\,dx\). Then \[ \int x\sin(4x^2-1)dx=\frac18\int\sin u\,du=\boxed{-\frac18\cos(4x^2-1)+C}. \]
- Let \(u=1+x^2\), so \(du=2x\,dx\). Then \[ \int\frac{x}{\sqrt{1+x^2}}dx=\frac12\int u^{-1/2}du=\boxed{\sqrt{1+x^2}+C}. \]
- Let \(u=\sqrt{x}\), so \(x=u^2\) and \(dx=2u\,du\). Then \[ \int\frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^u\,du=\boxed{2e^{\sqrt{x}}+C}. \]
Determine the value of each of the following limits. Show all of your work and justify your conclusions. Indicate if a limit approaches \(\displaystyle \infty\) or \(\displaystyle -\infty\) otherwise print DNE if the limit does not exist.
(a)
(b)
Solution:
- The expression \[ \lim_{x\to1}\frac{\ln x}{x-1} \] is the difference quotient for \(\ln x\) at \(x=1\), because \(\ln 1=0\). So this limit is exactly \[ (\ln x)'|_{x=1}. \] Since \(\dfrac{d}{dx}\ln x=\dfrac1x\), the value at \(x=1\) is \[ \boxed{1}. \]
- As \(x\to1^+\), the numerator tends to \[ 1^2-5(1)+5=1, \] which is positive. The denominator is \[ (x-1)(x+2). \] When \(x\to1^+\), we have \(x-1\to0^+\) and \(x+2\to3^+\), so the denominator tends to \(0^+\). Therefore the quotient is a positive number divided by something very small and positive, so it tends to \[ \boxed{+\infty}. \]
The following questions refer to the limit definition of the derivative.
- State the limit definition of the derivative of the function \(\displaystyle f(x)\) at the point \(\displaystyle x_0\).
- Use the \textbf{limit definition of the derivative} to show that the derivative of \(\displaystyle f(x) = 3x^2 - 4x + 10\) at \(\displaystyle x_0\) is \(\displaystyle f'(x_0) = 6x_0 - 4\).
Solution:
- The limit definition at \(x_0\) is \[ f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}, \] provided the limit exists.
- For \(f(x)=3x^2-4x+10\), \[ \frac{f(x_0+h)-f(x_0)}{h} =\frac{3(x_0+h)^2-4(x_0+h)+10-(3x_0^2-4x_0+10)}{h}. \] Expand the square and simplify: \[ \frac{3x_0^2+6x_0h+3h^2-4x_0-4h+10-3x_0^2+4x_0-10}{h} =\frac{6x_0h+3h^2-4h}{h}. \] Factor out \(h\): \[ \frac{h(6x_0+3h-4)}{h}=6x_0+3h-4. \] Now take the limit as \(h\to0\): \[ f'(x_0)=\lim_{h\to0}(6x_0+3h-4)=6x_0-4. \] Therefore \[ \boxed{f'(x_0)=6x_0-4}. \]
Use the axes below to make a sketch of the graph of a function whose domain is \(\displaystyle (-\infty, \infty)\) that satisfies all of the following criteria:
\(\displaystyle f(x)\) is \(\displaystyle \textbf{not}\) differentiable at \(\displaystyle x = 4\).
Sketch of a Function
[A set of axes with labeled gridlines for sketching the graph]
IMAGE
Solution: A full-credit sketch must satisfy each condition separately.
First, as \(x\to-\infty\), the graph should level off toward the horizontal line \(y=-2\). As \(x\to1^-\), the graph should approach the height \(4\), but because \(f(1)=1\), the actual plotted point at \(x=1\) must be the filled point \((1,1)\), not \((1,4)\).
At \(x=2\), the left-hand limit must be \(3\) and the right-hand limit must be \(5\), so the graph should approach height \(3\) from the left and height \(5\) from the right. Since \(f(2)=-1\), the actual point at \(x=2\) must be a filled dot at \((2,-1)\).
As \(x\to\infty\), the graph should level off toward the horizontal line \(y=1\). Finally, the function must fail to be differentiable at \(x=4\), so the sketch should include a corner, cusp, or vertical tangent there.
Any graph that satisfies all of these features simultaneously earns full credit.
Popeye and Olive Oyl will be sailing away from the same dock on the small island they call home. Popeye will sail directly East at 5 knots starting at 1pm. (One knot is one nautical mile per hour.) One hour later (2pm) Olive Oyl will leave the island sailing directly South at 4 knots. Determine the rate of change of the distance between them at 3pm.
Solution: Let \(x\) be Popeye's eastward distance, \(y\) be Olive's southward distance, and \(z\) be the distance between them.
At \(3\)pm, Popeye has been sailing for \(2\) hours at \(5\) knots, so \[ x=5(2)=10. \] Olive has been sailing for \(1\) hour at \(4\) knots, so \[ y=4(1)=4. \] These distances form the legs of a right triangle, so \[ z^2=x^2+y^2. \] Differentiate with respect to time: \[ 2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}. \] At that instant, \[ \frac{dx}{dt}=5,\qquad \frac{dy}{dt}=4, \] and \[ z=\sqrt{10^2+4^2}=\sqrt{116}=2\sqrt{29}. \] Substitute into the related-rates equation: \[ \frac{dz}{dt}=\frac{10\cdot5+4\cdot4}{2\sqrt{29}}=\frac{33}{\sqrt{29}}\approx\boxed{6.13\text{ knots}}. \] So the distance between Popeye and Olive is increasing at approximately \(6.13\) knots at \(3\)pm.
Determine the absolute minimum and absolute maximum of the function
over the interval \(\displaystyle -2 \leq x \leq 2\).
Solution: Since we are looking for absolute extrema on the closed interval \([-2,2]\), we find all critical points in the interval and compare the function values there and at the endpoints.
Write \[ f(x)=e^{-x}(x^2-3). \] Differentiate: \[ f'(x)=e^{-x}(2x)-(e^{-x})(x^2-3)=e^{-x}(-x^2+2x+3)=e^{-x}(3-x)(x+1). \] Because \(e^{-x}\neq0\), the critical points come from \[ (3-x)(x+1)=0. \] So the critical numbers are \(x=-1\) and \(x=3\), but only \(x=-1\) lies in \([-2,2]\).
Now evaluate \(f\) at the endpoints and at the interior critical point: \[ f(-2)=e^2,\qquad f(-1)=-2e,\qquad f(2)=e^{-2}. \] Among these values, the largest is \(e^2\) and the smallest is \(-2e\). Therefore the absolute maximum is \[ \boxed{e^2 \text{ at } x=-2}, \] and the absolute minimum is \[ \boxed{-2e \text{ at } x=-1}. \]
A function, \(\displaystyle f(x)\), and its derivative, \(\displaystyle f'(x)\), are both defined and continuous on the interval \(\displaystyle 0 \leq x \leq 4\). They satisfy the following criteria:
\(\displaystyle f'(0) = 0\), and \(\displaystyle f'(3) = 0\).
\(\displaystyle f'(x)\) is strictly increasing on the interval \(\displaystyle 0 \leq x < 2\).
\(\displaystyle f'(x)\) is strictly decreasing on the interval \(\displaystyle 2 < x \leq 4\).
\(\displaystyle f(0) = -2\).
- Provide a sketch of the graph \(\displaystyle f'(x)\) that is consistent with the conditions above.
Sketch of \(\displaystyle f'(x)\)
- Provide a sketch of the graph of \(\displaystyle f(x)\) that is consistent with your sketch above.
Sketch of \(\displaystyle f(x)\)
IMAGE
Solution:
- Because \(f'(0)=0\) and \(f'(3)=0\), the graph of \(f'\) must pass through the points \((0,0)\) and \((3,0)\). Since \(f'\) is strictly increasing on \(0\le x<2\), its graph must rise on that interval. Since \(f'\) is strictly decreasing on \(2<x\le4\), it must come down after \(x=2\). So a consistent sketch is a continuous curve that starts at \((0,0)\), rises to a peak near \(x=2\), and then decreases back through \((3,0)\).
- Now interpret this information for \(f\). We are given \(f(0)=-2\), so the graph of \(f\) must pass through \((0,-2)\). Where \(f'>0\), the function \(f\) is increasing; where \(f'\) is increasing, we have \(f''>0\), so \(f\) is concave up; and where \(f'\) is decreasing, we have \(f''<0\), so \(f\) is concave down. Thus a consistent sketch of \(f\) begins at \((0,-2)\), increases, is concave up on \((0,2)\), changes concavity near \(x=2\), and is concave down on \((2,4)\), with horizontal tangents at \(x=0\) and \(x=3\).
The graph of the derivative, \(\displaystyle f'(x)\), of a function is given below. Use the graph to answer each of the questions below. For each question provide a brief, one sentence justification for your answer.
Sketch of the Derivative of a Function
- Values of \(\displaystyle x\) where \(\displaystyle f(x)\) is increasing.
- Values of \(\displaystyle x\) where \(\displaystyle f(x)\) is decreasing.
- Value(s) of \(\displaystyle x\) where \(\displaystyle f(x)\) has a local maxima or minima.
Max: ________________
Min: ________________
IMAGE
Solution:
- \(f\) is increasing where \(f'(x)>0\). From the graph this occurs on \(\boxed{(0,4)\cup(5,6)}\), with the tangent horizontal at \(x=2\) but no sign change there.
- Similarly, \(f\) is decreasing where \(f'(x)<0\), namely on \(\boxed{(4,5)}\).
- A local maximum occurs where \(f'\) changes from positive to negative, so \(\boxed{x=4}\) is a local maximum. A local minimum occurs where \(f'\) changes from negative to positive, so \(\boxed{x=5}\) is a local minimum. The point \(x=2\) is not an extremum because the derivative only touches zero there.
Consider the region that is bounded on the left by \(\displaystyle x = 1\), bounded on the right by \(\displaystyle x = 4\), bounded above by \(\displaystyle f(x) = x^2\), and bounded below by \(\displaystyle g(x) = -x\).
- Make a sketch of the region.
- Construct an estimate of the area of the region using a Riemann sum with four rectangles of equal length. Do not provide the number but provide an expression that can be directly plugged into a calculator. Also, state if the sum is a left sided sum, right sided sum, or mid-point sum for your estimate.
- Generalize the estimate for the area to a Riemann sum with \(\displaystyle N\) rectangles of equal width.
IMAGE
Solution:
- A sketch should show the region between the curves \(y=x^2\) and \(y=-x\) from \(x=1\) to \(x=4\), with the parabola above the line throughout that interval.
- If we use four equal subintervals, then \(\Delta x=\dfrac{4-1}{4}=\dfrac34\). One acceptable estimate is the left-endpoint sum \[ \boxed{\frac34\sum_{k=0}^{3}\left(\left(1+\frac{3k}{4}\right)^2-\left(-1-\frac{3k}{4}\right)\right)}, \] which is a left-sided Riemann sum.
- With \(N\) equal rectangles, \(\Delta x=\dfrac{3}{N}\), and a left-endpoint sum is \[ \boxed{\frac{3}{N}\sum_{k=0}^{N-1}\left(\left(1+\frac{3k}{N}\right)^2+1+\frac{3k}{N}\right)}. \]
The velocity of a particle is
where the unit for \(\displaystyle t\) is seconds, and the units for the velocity are meters per second. Determine the displacement from \(\displaystyle t = 0\) to \(\displaystyle t = 60\).
Solution: The displacement is \[ \int_0^{60}\bigl(\sin(3\pi t)+e^{-t}+1\bigr)dt. \] Now \[ \int_0^{60}\sin(3\pi t)dt=0 \] because 60 seconds contains an integer number of periods, and \[ \int_0^{60}e^{-t}dt=1-e^{-60},\qquad \int_0^{60}1\,dt=60. \] Therefore the displacement is \[ \boxed{61-e^{-60}\text{ meters}}. \]
Two small vehicles are placed on opposite sides of a straight track facing each other. The track is 100 meters long. An experiment is conducted in which the cars move toward each other with constant velocities until they crash into one another. The cost of fuel for the first car is four times the square of the distance it moves. The cost of fuel for the second car is two times the square of the distance it moves. The experimenters want to minimize the fuel costs. How far should the first car travel? Justify that your answer is a minimum.
Solution: Let \(x\) be the distance traveled by the first car, so the second travels \(100-x\). The total fuel cost is \[ C(x)=4x^2+2(100-x)^2. \] Differentiate: \[ C'(x)=8x-4(100-x)=12x-400. \] Set \(C'(x)=0\): \(12x=400\), so \(x=\dfrac{100}{3}\). Since \(C''(x)=12>0\), this critical point gives a minimum. Thus the first car should travel \(\boxed{\dfrac{100}{3}\text{ meters}}\).
Consider the ellipse given by
- Determine the equation of the tangent line to the curve at the point \(\displaystyle (\sqrt{2}, 0)\).
- Determine all points on the ellipse which have a horizontal tangent line.
Solution:
- Differentiate implicitly: \[ 2x+2y+2xy'+6yy'=0. \] So \[ y'=-\frac{x+y}{x+3y}. \] At \((\sqrt2,0)\), the slope is \(-1\), so the tangent line is \[ \boxed{y=-x+\sqrt2}. \]
- A horizontal tangent requires \(y'=0\), so \(x+y=0\), hence \(y=-x\). Substitute into the ellipse: \[ x^2-2x^2+3x^2=2\implies2x^2=2\implies x=\pm1. \] Thus the points are \(\boxed{(1,-1)\text{ and }(-1,1)}\).
How well do you think your score will be on this test compared to the other students? Circle the percentage below that corresponds to your rank. A 100% means that you will have the best score, and a 50% means that your score will be in the middle of all of the scores.
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
out of a possible 12 points
Solution: This item is the self-ranking question rather than a calculus problem, so there is no mathematical solution to add here. The copied file keeps the prompt and simply notes that no worked solution is applicable.