Solved Practice Exercises 13-26. Implicit differentiation | Chegg.com - Free Printable
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Step-by-step solution for: Solved Practice Exercises 13-26. Implicit differentiation | Chegg.com
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Step-by-step solution for: Solved Practice Exercises 13-26. Implicit differentiation | Chegg.com
13. a. Differentiate implicitly: $4x^3 + 4y^3 \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x^3}{y^3}$.
b. At $(1, -1)$: $\frac{dy}{dx} = -\frac{1^3}{(-1)^3} = -\frac{1}{-1} = 1$.
14. a. Differentiate implicitly: $1 = e^y \frac{dy}{dx}$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = e^{-y}$.
b. At $(2, \ln 2)$: $\frac{dy}{dx} = e^{-\ln 2} = \frac{1}{2}$.
15. a. Differentiate implicitly: $2y \frac{dy}{dx} = 4$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2}{y}$.
b. At $(1, 2)$: $\frac{dy}{dx} = \frac{2}{2} = 1$.
16. a. Differentiate implicitly: $2y \frac{dy}{dx} + 3 = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{3}{2y}$.
b. At $(1, \sqrt{5})$: $\frac{dy}{dx} = -\frac{3}{2\sqrt{5}} = -\frac{3\sqrt{5}}{10}$.
17. a. Differentiate implicitly: $\cos y \frac{dy}{dx} = 20x^3$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{20x^3}{\cos y}$.
b. At $(1, \pi)$: $\frac{dy}{dx} = \frac{20(1)^3}{\cos \pi} = \frac{20}{-1} = -20$.
18. a. Differentiate implicitly: $\frac{1}{2\sqrt{x}} - 2 \cdot \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$. Simplify: $\frac{1}{2\sqrt{x}} - \frac{1}{\sqrt{y}} \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\sqrt{y}}{2\sqrt{x}}$.
b. At $(4, 1)$: $\frac{dy}{dx} = \frac{\sqrt{1}}{2\sqrt{4}} = \frac{1}{2 \cdot 2} = \frac{1}{4}$.
19. a. Differentiate implicitly: $-\sin y \frac{dy}{dx} = 1$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{1}{\sin y}$.
b. At $(0, \frac{\pi}{2})$: $\frac{dy}{dx} = -\frac{1}{\sin(\frac{\pi}{2})} = -\frac{1}{1} = -1$.
20. a. Differentiate implicitly: $\sec^2(xy) \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$. Solve for $\frac{dy}{dx}$: $\sec^2(xy) y + \sec^2(xy) x \frac{dy}{dx} = 1 + \frac{dy}{dx}$. Rearrange: $\sec^2(xy) x \frac{dy}{dx} - \frac{dy}{dx} = 1 - \sec^2(xy) y$. Factor: $\frac{dy}{dx} (\sec^2(xy) x - 1) = 1 - \sec^2(xy) y$. Solve: $\frac{dy}{dx} = \frac{1 - \sec^2(xy) y}{\sec^2(xy) x - 1}$.
b. At $(0, 0)$: $\sec^2(0) = 1$, so $\frac{dy}{dx} = \frac{1 - 1 \cdot 0}{1 \cdot 0 - 1} = \frac{1}{-1} = -1$.
21. a. Differentiate implicitly: $y + x \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{y}{x}$.
b. At $(1, 7)$: $\frac{dy}{dx} = -\frac{7}{1} = -7$.
22. a. Differentiate implicitly: $\frac{(y^2 + 1)(1) - x(2y \frac{dy}{dx})}{(y^2 + 1)^2} = 0$. Simplify: $y^2 + 1 - 2xy \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y^2 + 1}{2xy}$.
b. At $(10, 3)$: $\frac{dy}{dx} = \frac{3^2 + 1}{2 \cdot 10 \cdot 3} = \frac{10}{60} = \frac{1}{6}$.
23. a. Differentiate implicitly: $\frac{1}{3} x^{-2/3} + \frac{1}{3} y^{-2/3} \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x^{-2/3}}{y^{-2/3}} = -\left( \frac{y}{x} \right)^{2/3}$.
b. At $(1, 1)$: $\frac{dy}{dx} = -\left( \frac{1}{1} \right)^{2/3} = -1$.
24. a. Differentiate implicitly: $\frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left( \frac{y}{x} \right)^{1/3}$.
b. At $(1, 1)$: $\frac{dy}{dx} = -\left( \frac{1}{1} \right)^{1/3} = -1$.
25. a. Differentiate implicitly: $\sqrt[3]{y} + x \cdot \frac{1}{3} y^{-2/3} \frac{dy}{dx} + \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} \left( \frac{x}{3} y^{-2/3} + 1 \right) = -\sqrt[3]{y}$. So $\frac{dy}{dx} = -\frac{\sqrt[3]{y}}{\frac{x}{3} y^{-2/3} + 1}$.
b. At $(1, 8)$: $\sqrt[3]{8} = 2$, $y^{-2/3} = 8^{-2/3} = (2^3)^{-2/3} = 2^{-2} = \frac{1}{4}$. So $\frac{dy}{dx} = -\frac{2}{\frac{1}{3} \cdot \frac{1}{4} + 1} = -\frac{2}{\frac{1}{12} + 1} = -\frac{2}{\frac{13}{12}} = -2 \cdot \frac{12}{13} = -\frac{24}{13}$.
26. a. Differentiate implicitly: $\frac{2}{3} (x + y)^{-1/3} (1 + \frac{dy}{dx}) = \frac{dy}{dx}$. Solve for $\frac{dy}{dx}$: $\frac{2}{3} (x + y)^{-1/3} + \frac{2}{3} (x + y)^{-1/3} \frac{dy}{dx} = \frac{dy}{dx}$. Rearrange: $\frac{2}{3} (x + y)^{-1/3} = \frac{dy}{dx} - \frac{2}{3} (x + y)^{-1/3} \frac{dy}{dx} = \frac{dy}{dx} \left( 1 - \frac{2}{3} (x + y)^{-1/3} \right)$. Solve: $\frac{dy}{dx} = \frac{\frac{2}{3} (x + y)^{-1/3}}{1 - \frac{2}{3} (x + y)^{-1/3}}$.
b. At $(4, 4)$: $x + y = 8$, $(x + y)^{-1/3} = 8^{-1/3} = \frac{1}{2}$. So $\frac{dy}{dx} = \frac{\frac{2}{3} \cdot \frac{1}{2}}{1 - \frac{2}{3} \cdot \frac{1}{2}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$.
b. At $(1, -1)$: $\frac{dy}{dx} = -\frac{1^3}{(-1)^3} = -\frac{1}{-1} = 1$.
14. a. Differentiate implicitly: $1 = e^y \frac{dy}{dx}$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = e^{-y}$.
b. At $(2, \ln 2)$: $\frac{dy}{dx} = e^{-\ln 2} = \frac{1}{2}$.
15. a. Differentiate implicitly: $2y \frac{dy}{dx} = 4$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2}{y}$.
b. At $(1, 2)$: $\frac{dy}{dx} = \frac{2}{2} = 1$.
16. a. Differentiate implicitly: $2y \frac{dy}{dx} + 3 = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{3}{2y}$.
b. At $(1, \sqrt{5})$: $\frac{dy}{dx} = -\frac{3}{2\sqrt{5}} = -\frac{3\sqrt{5}}{10}$.
17. a. Differentiate implicitly: $\cos y \frac{dy}{dx} = 20x^3$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{20x^3}{\cos y}$.
b. At $(1, \pi)$: $\frac{dy}{dx} = \frac{20(1)^3}{\cos \pi} = \frac{20}{-1} = -20$.
18. a. Differentiate implicitly: $\frac{1}{2\sqrt{x}} - 2 \cdot \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$. Simplify: $\frac{1}{2\sqrt{x}} - \frac{1}{\sqrt{y}} \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\sqrt{y}}{2\sqrt{x}}$.
b. At $(4, 1)$: $\frac{dy}{dx} = \frac{\sqrt{1}}{2\sqrt{4}} = \frac{1}{2 \cdot 2} = \frac{1}{4}$.
19. a. Differentiate implicitly: $-\sin y \frac{dy}{dx} = 1$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{1}{\sin y}$.
b. At $(0, \frac{\pi}{2})$: $\frac{dy}{dx} = -\frac{1}{\sin(\frac{\pi}{2})} = -\frac{1}{1} = -1$.
20. a. Differentiate implicitly: $\sec^2(xy) \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$. Solve for $\frac{dy}{dx}$: $\sec^2(xy) y + \sec^2(xy) x \frac{dy}{dx} = 1 + \frac{dy}{dx}$. Rearrange: $\sec^2(xy) x \frac{dy}{dx} - \frac{dy}{dx} = 1 - \sec^2(xy) y$. Factor: $\frac{dy}{dx} (\sec^2(xy) x - 1) = 1 - \sec^2(xy) y$. Solve: $\frac{dy}{dx} = \frac{1 - \sec^2(xy) y}{\sec^2(xy) x - 1}$.
b. At $(0, 0)$: $\sec^2(0) = 1$, so $\frac{dy}{dx} = \frac{1 - 1 \cdot 0}{1 \cdot 0 - 1} = \frac{1}{-1} = -1$.
21. a. Differentiate implicitly: $y + x \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{y}{x}$.
b. At $(1, 7)$: $\frac{dy}{dx} = -\frac{7}{1} = -7$.
22. a. Differentiate implicitly: $\frac{(y^2 + 1)(1) - x(2y \frac{dy}{dx})}{(y^2 + 1)^2} = 0$. Simplify: $y^2 + 1 - 2xy \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y^2 + 1}{2xy}$.
b. At $(10, 3)$: $\frac{dy}{dx} = \frac{3^2 + 1}{2 \cdot 10 \cdot 3} = \frac{10}{60} = \frac{1}{6}$.
23. a. Differentiate implicitly: $\frac{1}{3} x^{-2/3} + \frac{1}{3} y^{-2/3} \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x^{-2/3}}{y^{-2/3}} = -\left( \frac{y}{x} \right)^{2/3}$.
b. At $(1, 1)$: $\frac{dy}{dx} = -\left( \frac{1}{1} \right)^{2/3} = -1$.
24. a. Differentiate implicitly: $\frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left( \frac{y}{x} \right)^{1/3}$.
b. At $(1, 1)$: $\frac{dy}{dx} = -\left( \frac{1}{1} \right)^{1/3} = -1$.
25. a. Differentiate implicitly: $\sqrt[3]{y} + x \cdot \frac{1}{3} y^{-2/3} \frac{dy}{dx} + \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} \left( \frac{x}{3} y^{-2/3} + 1 \right) = -\sqrt[3]{y}$. So $\frac{dy}{dx} = -\frac{\sqrt[3]{y}}{\frac{x}{3} y^{-2/3} + 1}$.
b. At $(1, 8)$: $\sqrt[3]{8} = 2$, $y^{-2/3} = 8^{-2/3} = (2^3)^{-2/3} = 2^{-2} = \frac{1}{4}$. So $\frac{dy}{dx} = -\frac{2}{\frac{1}{3} \cdot \frac{1}{4} + 1} = -\frac{2}{\frac{1}{12} + 1} = -\frac{2}{\frac{13}{12}} = -2 \cdot \frac{12}{13} = -\frac{24}{13}$.
26. a. Differentiate implicitly: $\frac{2}{3} (x + y)^{-1/3} (1 + \frac{dy}{dx}) = \frac{dy}{dx}$. Solve for $\frac{dy}{dx}$: $\frac{2}{3} (x + y)^{-1/3} + \frac{2}{3} (x + y)^{-1/3} \frac{dy}{dx} = \frac{dy}{dx}$. Rearrange: $\frac{2}{3} (x + y)^{-1/3} = \frac{dy}{dx} - \frac{2}{3} (x + y)^{-1/3} \frac{dy}{dx} = \frac{dy}{dx} \left( 1 - \frac{2}{3} (x + y)^{-1/3} \right)$. Solve: $\frac{dy}{dx} = \frac{\frac{2}{3} (x + y)^{-1/3}}{1 - \frac{2}{3} (x + y)^{-1/3}}$.
b. At $(4, 4)$: $x + y = 8$, $(x + y)^{-1/3} = 8^{-1/3} = \frac{1}{2}$. So $\frac{dy}{dx} = \frac{\frac{2}{3} \cdot \frac{1}{2}}{1 - \frac{2}{3} \cdot \frac{1}{2}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$.
Parent Tip: Review the logic above to help your child master the concept of implicit differentiation worksheet.
