Product Rule - Calculus 1

1. What is the value of $\frac{dy}{dx}$ at the point $(2, \sqrt{2})$?
   Answer: $-\frac{3\sqrt{2}}{4}$ We use implicit differentiation with respect to $x$. The left side, $x^3y^2$, requires the Product Rule. $\frac{d}{dx}(x^3 y^2) = \frac{d}{dx}(8)$ $(\frac{d}{dx}x^3)y^2 + x^3(\frac{d}{dx}y^2) = 0$ $(3x^2)y^2 + x^3(2y \frac{dy}{dx}) = 0$. Now, solve for $\frac{dy}{dx}$: $2x^3y \frac{dy}{dx} = -3x^2y^2$ $\frac{dy}{dx} = \frac{-3x^2y^2}{2x^3y} = \frac{-3y}{2x}$. Finally, substitute the coordinates of the point $(2, \sqrt{2})$ into this expression: $\frac{dy}{dx} = \frac{-3(\sqrt{2})}{2(2)} = -\frac{3\sqrt{2}}{4}$.
2. What is the equation of the tangent line to the graph of $h(x) = x^2 \ln(x)$ at the point where $x=e$?
   Answer: $y = 3ex - 2e^2$ First, find the point of tangency by evaluating $h(e)$: $h(e) = e^2 \ln(e) = e^2(1) = e^2$. The point is $(e, e^2)$. Next, find the slope of the tangent line by finding the derivative $h'(x)$ using the Product Rule. Let $f(x) = x^2$ and $g(x) = \ln(x)$. $h'(x) = f'(x)g(x) + f(x)g'(x) = (2x)(\ln(x)) + (x^2)(\frac{1}{x}) = 2x\ln(x) + x$. Now, evaluate the slope at $x=e$: $h'(e) = 2e\ln(e) + e = 2e(1) + e = 3e$. The equation of the tangent line in point-slope form is $y - y_1 = m(x - x_1)$: $y - e^2 = 3e(x - e)$ $y - e^2 = 3ex - 3e^2$ $y = 3ex - 2e^2$.
3. Let $f(x) = x^2 g(x)$. If $g(2) = 3$ and $f'(2) = 8$, what is the value of $g'(2)$?
   Answer: $-1$ First, find the derivative of $f(x)$ using the product rule: $f'(x) = (\frac{d}{dx}x^2)g(x) + x^2 g'(x) = 2x g(x) + x^2 g'(x)$. Now, substitute $x=2$ into the equation: $f'(2) = 2(2)g(2) + (2)^2 g'(2) = 4g(2) + 4g'(2)$. We are given $f'(2)=8$ and $g(2)=3$. Substitute these values: $8 = 4(3) + 4g'(2)$. This simplifies to $8 = 12 + 4g'(2)$. Subtracting 12 from both sides gives $-4 = 4g'(2)$. Finally, dividing by 4 gives $g'(2) = -1$. * A is incorrect because it represents the intermediate value $4g'(2)$ before dividing by 4. * B is incorrect and would result from neglecting the first term of the product rule, assuming $f'(x) = x^2 g'(x)$, which leads to $8 = 4g'(2)$ or $g'(2)=2$. * C is incorrect due to a sign error in the final step of the algebra.
4. Which of the following is the derivative of $h(x) = x e^x \cos(x)$?
   Answer: $e^x(\cos(x) + x\cos(x) - x\sin(x))$ To find the derivative of a product of three functions, we can apply the Product Rule twice. Let's group $f(x) = (xe^x)$ and $g(x) = \cos(x)$. $h'(x) = f'(x)g(x) + f(x)g'(x)$. First, find $f'(x)$ using the Product Rule: $f'(x) = (1)e^x + x(e^x) = e^x + xe^x$. Now substitute this and $g'(x) = -\sin(x)$ into the main formula: $h'(x) = (e^x + xe^x)\cos(x) + (xe^x)(-\sin(x))$ $h'(x) = e^x\cos(x) + xe^x\cos(x) - xe^x\sin(x)$ Factoring out $e^x$ gives: $h'(x) = e^x(\cos(x) + x\cos(x) - x\sin(x))$.
5. What is the equation of the line tangent to the curve $y = x^2 \cos(x)$ at $x = \pi$?
   Answer: $y = -2\pi x + \pi^2$ First, find the y-coordinate of the point of tangency: $y(\pi) = \pi^2 \cos(\pi) = \pi^2(-1) = -\pi^2$. The point is $(\pi, -\pi^2)$. Next, find the derivative using the product rule: $y' = \frac{d}{dx}(x^2)\cos(x) + x^2\frac{d}{dx}(\cos(x)) = 2x\cos(x) - x^2\sin(x)$. Evaluate the derivative at $x=\pi$ to find the slope of the tangent line: $y'(\pi) = 2\pi\cos(\pi) - \pi^2\sin(\pi) = 2\pi(-1) - \pi^2(0) = -2\pi$. Finally, use the point-slope form: $y - y_1 = m(x - x_1)$, which gives $y - (-\pi^2) = -2\pi(x - \pi)$. Simplifying, we get $y + \pi^2 = -2\pi x + 2\pi^2$, so $y = -2\pi x + \pi^2$. * A is incorrect because it uses a slope of 0, which would happen if the derivative was incorrectly calculated as $y' = 2x(-\sin(x))$. * C is incorrect due to a sign error when substituting the y-coordinate into the point-slope formula. * D is incorrect due to an algebraic error in the final step of isolating y: adding $\pi^2$ to both sides instead of subtracting.
6. If $f(x) = \sqrt{x} \ln(x)$, what is the value of $f'(e^2)$?
   Answer: $\frac{2}{e}$ First, find the derivative of $f(x) = x^{1/2} \ln(x)$ using the product rule. $f'(x) = (\frac{d}{dx}x^{1/2})\ln(x) + x^{1/2}(\frac{d}{dx}\ln(x)) = (\frac{1}{2}x^{-1/2})\ln(x) + x^{1/2}(\frac{1}{x})$. This simplifies to $f'(x) = \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}} = \frac{\ln(x)+2}{2\sqrt{x}}$. Now, evaluate at $x=e^2$: $f'(e^2) = \frac{\ln(e^2)+2}{2\sqrt{e^2}} = \frac{2+2}{2e} = \frac{4}{2e} = \frac{2}{e}$. * A is incorrect if the second term of the product rule, which contributes the '+2' in the numerator, is forgotten. * C is incorrect if the property $\ln(e^2)=2$ is incorrectly recalled as $\ln(e^2)=1$. * D is incorrect due to a significant calculation error, possibly involving misplacing $e$ in the numerator.
7. If $f(x) = \sin(2x)\cos(x)$, what is the value of $f'(\frac{\pi}{6})$?
   Answer: $\frac{\sqrt{3}}{4}$ We use the Product Rule, $f'(x) = u'v + uv'$, combined with the Chain Rule for the term $\sin(2x)$. Let $u(x) = \sin(2x)$ and $v(x) = \cos(x)$. $u'(x) = \cos(2x) \cdot 2 = 2\cos(2x)$. $v'(x) = -\sin(x)$. So, $f'(x) = (2\cos(2x))(\cos(x)) + (\sin(2x))(-\sin(x)) = 2\cos(2x)\cos(x) - \sin(2x)\sin(x)$. Now we evaluate at $x = \frac{\pi}{6}$: $f'(\frac{\pi}{6}) = 2\cos(2 \cdot \frac{\pi}{6})\cos(\frac{\pi}{6}) - \sin(2 \cdot \frac{\pi}{6})\sin(\frac{\pi}{6})$ $= 2\cos(\frac{\pi}{3})\cos(\frac{\pi}{6}) - \sin(\frac{\pi}{3})\sin(\frac{\pi}{6})$ Using the values $\cos(\frac{\pi}{3})=\frac{1}{2}$, $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$, $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$, and $\sin(\frac{\pi}{6})=\frac{1}{2}$: $= 2(\frac{1}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{3}}{2})(\frac{1}{2})$ $= \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}$.
8. Let $h(x) = x \ln(x)$. Find the second derivative, $h''(x)$.
   Answer: $\frac{1}{x}$ First, find the first derivative using the product rule: $h'(x) = (\frac{d}{dx}x)\ln(x) + x(\frac{d}{dx}\ln(x)) = (1)\ln(x) + x(\frac{1}{x}) = \ln(x) + 1$. Now, differentiate $h'(x)$ to find the second derivative: $h''(x) = \frac{d}{dx}(\ln(x) + 1) = \frac{1}{x} + 0 = \frac{1}{x}$. * A is incorrect because this is the first derivative, $h'(x)$, not the second. * C is incorrect because it would be the result of an incorrect first derivative, such as if one made the common product rule error $h'(x) = (1)(\frac{1}{x}) = \frac{1}{x}$, and then differentiated again. * D is incorrect because it could result from incorrectly assuming the derivative of $\ln(x)+1$ is $1-1=0$.
9. For which values of $x$ does the function $f(x) = x^2 e^{-x}$ have a horizontal tangent line?
   Answer: $x=0, x=2$ A horizontal tangent line occurs where the derivative is zero. First, find the derivative using the product rule: $f'(x) = (\frac{d}{dx}x^2)e^{-x} + x^2(\frac{d}{dx}e^{-x}) = 2xe^{-x} + x^2(-e^{-x})$. Factor the expression: $f'(x) = xe^{-x}(2-x)$. Set the derivative equal to zero: $xe^{-x}(2-x) = 0$. Since $e^{-x}$ is never zero, we solve $x=0$ or $2-x=0$. This gives the solutions $x=0$ and $x=2$. * A is incorrect because it is an incomplete solution, missing the second value. * B is incorrect because it is an incomplete solution, missing the first value. * D is incorrect due to a sign error when solving the equation $2-x=0$.
10. Let $f(x)$ and $g(x)$ be differentiable functions. Let $p(x) = f(x)g(x)$ and $q(x) = \frac{f(x)}{g(x)}$. If $f(a) = 0$, $g(a) \neq 0$, and $f'(a) \neq 0$, what is the relationship between $p'(a)$ and $q'(a)$?
    Answer: $p'(a) = [g(a)]^2 q'(a)$ First, find an expression for $p'(a)$ using the product rule: $p'(x) = f'(x)g(x) + f(x)g'(x)$. At $x=a$, $p'(a) = f'(a)g(a) + f(a)g'(a)$. Since $f(a)=0$, this simplifies to $p'(a) = f'(a)g(a)$. Next, find an expression for $q'(a)$ using the quotient rule: $q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$. At $x=a$, $q'(a) = \frac{f'(a)g(a) - f(a)g'(a)}{[g(a)]^2}$. Since $f(a)=0$, this simplifies to $q'(a) = \frac{f'(a)g(a)}{[g(a)]^2}$. From this, we see that $f'(a)g(a) = [g(a)]^2 q'(a)$. Since $p'(a) = f'(a)g(a)$, we can conclude that $p'(a) = [g(a)]^2 q'(a)$. * A is incorrect. This would only be true if $[g(a)]^2 = 1$. * B is incorrect as there is no basis for a negative relationship. * D is incorrect as it is missing a factor of $g(a)$.
11. To find the derivative of $y = (x^2+1)^{\sin(x)}$ using logarithmic differentiation, one must differentiate the expression $\sin(x) \ln(x^2+1)$. What is the result of this differentiation?
    Answer: $\cos(x) \ln(x^2+1) + \frac{2x \sin(x)}{x^2+1}$ We must use the product rule on $\sin(x) \ln(x^2+1)$. Let $f(x) = \sin(x)$ and $g(x) = \ln(x^2+1)$. Then $f'(x) = \cos(x)$ and, using the chain rule, $g'(x) = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}$. The product rule $f'g + fg'$ gives: $\cos(x) \ln(x^2+1) + \sin(x) \left(\frac{2x}{x^2+1}\right)$. * A is incorrect because it is the result of the common error of multiplying the derivatives, $f'g'$. * B is incorrect because it omits the factor of $2x$ from the chain rule when differentiating $\ln(x^2+1)$. * D is incorrect due to an incorrect sign on the second term.
12. Find the derivative of $h(x) = x e^x \sin(x)$.
    Answer: $e^x(\sin(x) + x\sin(x) + x\cos(x))$ The derivative of a product of three functions $(fgh)'$ is $f'gh + fg'h + fgh'$. Let $f(x)=x$, $g(x)=e^x$, and $h(x)=\sin(x)$. Their derivatives are $f'(x)=1$, $g'(x)=e^x$, and $h'(x)=\cos(x)$. Applying the rule: $h'(x) = (1)(e^x)(\sin(x)) + (x)(e^x)(\sin(x)) + (x)(e^x)(\cos(x))$. Factoring out $e^x$ gives $e^x(\sin(x) + x\sin(x) + x\cos(x))$. * A is incorrect because it is missing the term $fg'h = x e^x \sin(x)$. * B is incorrect as it results from the invalid rule $(fgh)' = f'g'h'$, which would be $(1)(e^x)(\cos x)$. * D is incorrect because it arises from a misunderstanding of how to combine the terms after differentiation.
13. If $f(x) = e^{2x} \sin(3x)$, what is $f'(x)$?
    Answer: $e^{2x}(2\sin(3x) + 3\cos(3x))$ This requires both the product rule and the chain rule. Let $u = e^{2x}$ and $v = \sin(3x)$. Then $u' = 2e^{2x}$ and $v' = 3\cos(3x)$. Applying the product rule, $f'(x) = u'v + uv'$, we get $f'(x) = (2e^{2x})\sin(3x) + (e^{2x})(3\cos(3x))$. Factoring out $e^{2x}$ gives $e^{2x}(2\sin(3x) + 3\cos(3x))$. * A is incorrect because it comes from the common misconception that the derivative of a product is the product of the derivatives: $u'v' = (2e^{2x})(3\cos(3x))$. * B is incorrect because it omits the factors of 2 and 3 that arise from the chain rule. * C is incorrect due to a sign error, as if the derivative of $\sin(3x)$ were $-3\cos(3x)$.
14. If $f(1) = 2$ and $f'(1) = -4$, what is the value of $k'(1)$?
    Answer: $0$ To find $k'(x)$, we use the Product Rule: $(uv)' = u'v + uv'$. Let $u(x) = 3x^2$ and $v(x) = f(x)$. Then $u'(x) = 6x$ and $v'(x) = f'(x)$. Applying the rule: $k'(x) = (6x)f(x) + (3x^2)f'(x)$. Now, substitute $x=1$ and the given values for $f(1)$ and $f'(1)$: $k'(1) = (6 \cdot 1)f(1) + (3 \cdot 1^2)f'(1)$ $k'(1) = 6(2) + 3(-4)$ $k'(1) = 12 - 12 = 0$.
15. Let $h(x) = f(x)g(x)$. If $f(2) = 3$, $f'(2) = -1$, $g(2) = 5$, and $g'(2) = 4$, what is the value of $h'(2)$?
    Answer: $7$ The product rule states that $h'(x) = f'(x)g(x) + f(x)g'(x)$. To find $h'(2)$, we substitute the given values: $h'(2) = f'(2)g(2) + f(2)g'(2) = (-1)(5) + (3)(4) = -5 + 12 = 7$. * A is incorrect because it results from the common mistake of multiplying the derivatives: $f'(2)g'(2) = (-1)(4) = -4$. * C is incorrect because it incorrectly uses the structure of the quotient rule: $f(2)g'(2) - f'(2)g(2) = (3)(4) - (-1)(5) = 12 + 5 = 17$. * D is incorrect because it incorrectly uses the structure of the quotient rule with a sign error: $f'(2)g(2) - f(2)g'(2) = (-1)(5) - (3)(4) = -5 - 12 = -17$.
16. If $h(x) = f(x)g(x)$ and $f'(x)g(x) = -f(x)g'(x)$ for all $x$, what must be true about $h(x)$?
    Answer: $h(x)$ is a constant function. The derivative of $h(x)$ is given by the product rule: $h'(x) = f'(x)g(x) + f(x)g'(x)$. The problem states that $f'(x)g(x) = -f(x)g'(x)$. We can rearrange this to $f'(x)g(x) + f(x)g'(x) = 0$. The left side of this equation is exactly the expression for $h'(x)$. Therefore, $h'(x) = 0$ for all $x$. If the derivative of a function is zero for all $x$, the function itself must be a constant. Thus, $h(x) = C$ for some constant $C$. * A is incorrect because the constant could be negative or zero. * B is incorrect because the constant could be positive or zero. * D is incorrect because an increasing function must have a non-negative derivative, and it cannot be zero everywhere unless it is constant.
17. Which of the following sets of conditions is sufficient to guarantee that $p'(c) > 0$ for some number $c$?
    Answer: $f(c) < 0, g(c) < 0, f'(c) < 0, g'(c) < 0$ The derivative of $p(x)$ is given by the Product Rule: $p'(x) = f'(x)g(x) + f(x)g'(x)$. We need to find the conditions that guarantee $p'(c) > 0$. Let's analyze each choice: A) $p'(c) = f'(c)g(c) + f(c)g'(c) = (\text{negative})(\text{positive}) + (\text{positive})(\text{positive}) = (\text{negative}) + (\text{positive})$. The sign is indeterminate. B) $p'(c) = f'(c)g(c) + f(c)g'(c) = (\text{negative})(\text{positive}) + (\text{negative})(\text{positive}) = (\text{negative}) + (\text{negative})$. This is always negative. C) $p'(c) = f'(c)g(c) + f(c)g'(c) = (\text{negative})(\text{negative}) + (\text{negative})(\text{negative}) = (\text{positive}) + (\text{positive})$. This is always positive. D) $p'(c) = f'(c)g(c) + f(c)g'(c) = (\text{positive})(\text{negative}) + (\text{positive})(\text{negative}) = (\text{negative}) + (\text{negative})$. This is always negative.
18. The line tangent to the graph of $f(x)$ at $x=2$ is $y = -x+5$, and the line tangent to the graph of $g(x)$ at $x=2$ is $y=3x-4$. If $p(x) = f(x)g(x)$, find the slope of the line tangent to the graph of $p(x)$ at $x=2$.
    Answer: $7$ The slope of the tangent line to $p(x)$ at $x=2$ is $p'(2)$. From the tangent line information, we can deduce the values of the functions and their derivatives at $x=2$. For $f(x)$, the slope is $f'(2) = -1$ and the point is $(2, -2+5) = (2,3)$, so $f(2)=3$. For $g(x)$, the slope is $g'(2) = 3$ and the point is $(2, 3(2)-4) = (2,2)$, so $g(2)=2$. Now, use the product rule for $p(x)$: $p'(x) = f'(x)g(x) + f(x)g'(x)$. At $x=2$, $p'(2) = f'(2)g(2) + f(2)g'(2) = (-1)(2) + (3)(3) = -2 + 9 = 7$. * A is incorrect because it is the product of the slopes, $f'(2)g'(2)$, a common error. * B is incorrect as it may result from misassigning the function and derivative values, for example, $f'(2)f(2) + g'(2)g(2) = (-1)(3)+(3)(2) = 3$. * D is incorrect due to a sign error on $f'(2)$: $(1)(2) + (3)(3) = 11$.
19. If $g(2) = 1$ and $g'(2) = 4$, what is the value of $h'(2)$?
    Answer: $-\frac{7e^2}{4}$ This problem tests your ability to apply both the quotient rule and product rule in combination. When you see a function that's a quotient where both numerator and denominator need differentiation, the quotient rule is your go-to tool. To find $h'(2)$, you need to differentiate $h(x) = \frac{e^x}{f(x)}$ using the quotient rule: $h'(x) = \frac{f(x) \cdot (e^x)' - e^x \cdot f'(x)}{[f(x)]^2} = \frac{f(x) \cdot e^x - e^x \cdot f'(x)}{[f(x)]^2}$ Since $f(x) = x g(x)$, you need $f'(x)$ using the product rule: $f'(x) = 1 \cdot g(x) + x \cdot g'(x) = g(x) + x g'(x)$ At $x = 2$: $f(2) = 2 \cdot g(2) = 2 \cdot 1 = 2$ and $f'(2) = g(2) + 2 \cdot g'(2) = 1 + 2 \cdot 4 = 9$ Therefore: $h'(2) = \frac{2 \cdot e^2 - e^2 \cdot 9}{2^2} = \frac{2e^2 - 9e^2}{4} = \frac{-7e^2}{4}$ Answer B is correct. Answer A represents forgetting the product rule when differentiating $f(x)$, using only $f'(2) = g(2) = 1$. Answer C likely comes from sign errors or incorrectly applying the quotient rule formula. Answer D results from reversing the quotient rule (putting $e^x \cdot f'(x) - f(x) \cdot e^x$ in the numerator instead). Remember: quotient rule problems often involve multiple differentiation rules. Work systematically through each component before substituting values, and double-check the quotient rule formula—the order matters for the sign.
20. In which step did the student first make an error?
    Answer: Step 3 The student makes their first error in Step 3. The product rule is $h'(x) = f'(x)g(x) + f(x)g'(x)$, not $h'(x) = f'(x)g'(x)$. This is a very common misconception. Steps 1 and 2 are correct identifications and differentiations of the component functions. Step 4 is a correct calculation based on the incorrect formula from Step 3, so the error originates in Step 3. * A is incorrect because Step 1 correctly identifies the functions in the product. * B is incorrect because Step 2 correctly finds the derivatives of the functions from Step 1. * D is incorrect because while the final answer is wrong, the error was made in the previous step's formula, not in the calculation in Step 4.

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