Practice Problems

In problems 1-19 find $\mathrm{dy} / \mathrm{dx}$ in two ways: (a) by differentiating implicitly and (b) by explicitly solving for $\mathrm{y}$ and then differentiating. Then find the value of $\mathrm{dy} / \mathrm{dx}$ at the given point using your results from both the implicit and the explicit differentiation.

1. $x^{2}+y^{2}=100$, point $(6,8)$

3. $x^{2}-3 x y+7 y=5$, point $(2,1)$

5. $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$, point $(0,4)$

7. $\ln (\mathrm{y})+3 \mathrm{x}-7=0$, point $(2, \mathrm{e})$

9. $x^{2}-y^{2}=16$, point $(5,-3)$

11. Find the slopes of the lines tangent to the graph in Fig. 3 at the points $(3,1),(3,3)$, and $(4,2)$.

13. Find the slopes of the lines tangent to the graph in Fig. 4 at the points $((5,0),(5,6)$, and $(-4,3)$.

In problems 15 – 21, find $\mathrm{dy} / \mathrm{dx}$ using implicit differentiation and then find the slope of the line tangent to the graph of the equation at the given point.

15. $y^{3}-5 y=5 x^{2}+7$, point $(1,3)$

17. $y^{2}+\sin (y)=2 x-6$, point $(3,0)$

19. $\mathrm{e}^{\mathrm{y}}+\sin (\mathrm{y})=\mathrm{x}^{2}-3$, point $(2,0)$

21. $x^{2 / 3}+y^{2 / 3}=5$, point $(8,1)$

23. Find the slope of the line tangent to the ellipse in Fig. 5 at the point $(1,2)$.

25. Find $\mathrm{y}^{\prime}$ for $y=A x^{2}+B x+C$ and for $x=A y^{2}+B y+C$.

27. Find $\mathrm{y}^{\prime}$ for $A x^{2}+B x y+C y^{2}+D x+E y+F=0$.

29. Find the coordinates of point A where the tangent line to the ellipse in Fig. 5 is horizontal.

31. Find the coordinates of points C and D on the ellipse in Fig. 5.

In problems 33-39 find $\mathrm{dy} / \mathrm{dx}$ in two ways: (a) by using the "usual" differentiation patterns and (b) by using logarithmic differentiation.

33. $y=\left(x^{2}+5\right)^{7} \cdot\left(x^{3}-1\right)^{4}$

35. $y=x^{5} \cdot(3 x+2)^{4}$

37. $y=e^{\sin (x)}$

39. $y=\sqrt{25-x^{2}}$

In problems 41–47, use logarithmic differentiation to find $\mathrm{dy} / \mathrm{dx}$.

41. $y=x^{\cos (x)}$

43. $y=x^{4} \cdot(x-2)^{7} \cdot \sin (3 x)$

45. $y=(3+\sin (x))^{x}$

In problems 47-49, use the values in each table to calculate the values of the derivative in the last column.

47. Use Table 1.

Table 1

$\begin{array}{c|c|c|c|c} \mathrm{x} & {\mathrm{f}(\mathrm{x})} & \ln (\mathrm{f}(\mathrm{x})) & \mathrm{D}(\ln (\mathrm{f}(\mathrm{x}))) & \mathrm{f}^{\prime}(\mathrm{x}) \\ \hline 1 & 1 & 0 & 1.2 & \\ 2 & 9 & 2.2 & 1.8 & \\ 3 & 64 & 4.2 & 2.1 & \end{array}$

49. Use Table 3.

Table 3

$\begin{array}{c|c|c|c|c} x & {\mathrm{f}(\mathrm{x})} & \ln (\mathrm{f}(\mathrm{x})) & \mathrm{D}(\ln (\mathrm{f}(\mathrm{x}))) & \mathrm{f}^{\prime}(\mathrm{x}) \\ \hline 1 & 5 & 1.6 & -1 & \\ 2 & 2 & 0.7 & 0 & \\ 3 & 7 & 1.9 & 2 & \end{array}$

Problems 51–55 illustrate how logarithmic differentiation can be used to verify some differentiation patterns we already know (51 and 52) and to derive some new patterns (53 – 55). Assume that all of the functions are differentiable and that the function combinations are defined.

51. Use logarithmic differentiation on $f \cdot g$ to rederive the product rule: $D(f \cdot g)=f \cdot g^{\prime}+g \cdot f^{\prime}$.

53. Use logarithmic differentiation on $\mathrm{f} \cdot \mathrm{g} \cdot \mathrm{h}$ to derive a product rule for three functions: $\mathbf{D}(\mathrm{f} \cdot \mathrm{g} \cdot \mathrm{h})$.

55. Use logarithmic differentiation to determine a pattern for the derivative of $\mathrm{f}^{\mathrm{g}}: \mathrm{D}\left(\mathrm{f}^{\mathrm{g}}\right)$.

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-3.10-Implicit-and-Logarithmic-Differentiation.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

 1. (a) $x^{2}+y^{2}=100$ so $2 x+2 y y^{\prime}=0$ and $y^{\prime}=-x / y.$ At $(6,8), y^{\prime}=-6 / 8=-3 / 4$.

    (b) $y=\sqrt{100-x^{2}}$ so $y^{\prime}=\frac{-x}{\sqrt{100-x^{2}}}.$ At $(6,8), y^{\prime}=\frac{-6}{\sqrt{100-36}}=\frac{-6}{8}=-\frac{3}{4}$.

3. (a) $x^{2}-3 x y+7 y=5$ so $2 x-3\left(y+x y^{\prime}\right)+7 y^{\prime}=0 \quad$ and $y^{\prime}=\frac{3 y-2 x}{7-3 x}$. At $(2,1), y^{\prime}=\frac{3-4}{7-6}=-1$.

    (b) $y=\frac{5-x^{2}}{7-3 x}$ so $y^{\prime}=\frac{(7-3 x)(-2 x)-\left(5-x^{2}\right)(-3)}{(7-3 x)^{2}}$. At $(2,1), y^{\prime}=\frac{(1)(-4)}{}$ At $(2,1), \mathrm{y}^{\prime} = \frac{(1)(-4)-(1)(-3)}{(1)^{2}}=-1$.

5. (a) $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ so $\frac{2 x}{9}+\frac{2 y}{16} \mathbf{y}^{\prime}=0$ and $\mathbf{y}^{\prime}=-\frac{16}{9} \frac{x}{y}.$ At $(0,4), y^{\prime}=0$.

    (b) $\mathrm{y}=4 \sqrt{1-\frac{\mathrm{x}^{2}}{9}}=\frac{4}{3} \sqrt{9-\mathrm{x}^{2}}$ so $\mathrm{y}^{\prime}=\frac{4}{3} \frac{-\mathrm{x}}{\sqrt{9-\mathrm{x}^{2}}}$. At $(0,4), \mathrm{y}^{\prime}=0$.

7. (a) $\ln (\mathrm{y})+3 \mathrm{x}-7=0$ so $\frac{1}{\mathrm{y}} \mathbf{y}^{\prime}+3=0$ and $\mathbf{y}^{\prime}=-3 \mathrm{y}.$ At $(2, \mathrm{e}), \mathrm{y}^{\prime}=-3 \mathrm{e}$.

    (b) $y=e^{7-3 x}$ so $y^{\prime}=-3 e^{7-3 x}.$ At $(2, e), y^{\prime}=-3 e^{7-6}=-3 e$.

9. (a) $x^{2}-y^{2}=16$ so $2 x-2 y y^{\prime}=0$ and $y^{\prime}=x / y$. At $(5,-3), y^{\prime}=-5 / 3$.

    (b) The point $(5,-3)$ is on the bottom half of the circle so

     $\mathrm{y}=-\sqrt{\mathrm{x}^{2}-16}.$ Then $\mathrm{y}^{\prime}=-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}-16}}.$ At $(5,-3), \mathrm{y}^{\prime}=-\frac{5}{\sqrt{25-16}}=-\frac{5}{3}$.

11. $x=4 y-y^{2}$ so, differentiating each side, $1=4 \mathbf{y}^{\prime}-2 y \mathbf{y}^{\prime}$ and $\mathbf{y}^{\prime}=\frac{1}{4-2 y}$.

      At $(3,1), \mathrm{y}^{\prime}=\frac{1}{4-2(1)}=\frac{1}{2}.$ At $(3,3), \mathrm{y}^{\prime}=\frac{1}{4-2(3)}=-\frac{1}{2}$.

       At $(4,2), y^{\prime}=\frac{1}{4-2(2)}$ is undefined (the tangent line is vertical).

13. $x=y^{2}-6 y+5$. Differentiating each side, $1=2 y y^{\prime}-6 y^{\prime}$ and $y^{\prime}=\frac{1}{2 y-6}$. At $(5,0), \mathrm{y}^{\prime}=\frac{1}{2(0)-6}=-\frac{1}{6}.$ At $(5,6), \mathrm{y}^{\prime}=\frac{1}{2(6)-6}=\frac{1}{6}$.

       At $(-4,3), \mathrm{y}^{\prime}=\frac{1}{2(3)-6}$ is undefined (vertical tangent line).

15. $3 y^{2} y^{\prime}-5 \mathrm{y}^{\prime}=10 x$ so $\mathrm{y}^{\prime}=\frac{10 x}{3 y^{2}-5}$. At $(1,3), m=10 / 22=5 / 11$.

17. $\mathrm{y}^{2}+\sin (\mathrm{y})=2 \mathrm{x}-6$ so $2 \mathrm{y} \mathrm{y}^{\prime}+\cos (\mathrm{y}) \mathbf{y}^{\prime}=2$ and $\mathbf{y}^{\prime}=\frac{2}{2 \mathrm{y}+\cos (\mathrm{y})}.$ At $(3,0), \mathrm{m}=\frac{2}{0+1}=2$.

19. $\mathrm{e}^{\mathrm{y}}+\sin (\mathrm{y})=\mathrm{x}^{2}-3$ so $\mathrm{e}^{\mathrm{y}} \mathbf{y}^{\prime}+\cos (\mathrm{y}) \mathbf{y}^{\prime}=2 \mathrm{x}$ and $\mathbf{y}^{\prime}=\frac{2 \mathrm{x}}{\mathrm{e}^{\mathrm{y}}+\cos (\mathrm{y})}$. At $(2,0), \mathrm{m}=\frac{4}{1+1}=2$.

21. $x^{2 / 3}+y^{2 / 3}=5$ so $\frac{2}{3} x^{-1 / 3}+\frac{2}{3} y^{-1 / 3} y^{\prime}=0$ and $y^{\prime}=-(x / y)^{-1 / 3}.$ At $(8,1), m=-(8 / 1)^{-1 / 3}=-\frac{1}{2}$

23. Using implicit differentiation, $2 \mathrm{x}+\mathrm{xy}^{\prime}+\mathrm{y}+2 \mathrm{y} \mathbf{y}^{\prime}+3-7 \mathbf{y}^{\prime}=0$ so $\mathbf{y}^{\prime}=\frac{-\mathbf{2} \mathbf{x}-\mathbf{y}-\mathbf{3}}{\mathbf{x}+\mathbf{2} \mathbf{y}-\mathbf{7}}$.

      At $(1,2), y^{\prime}=\frac{-2(1)-(2)-3}{(1)+2(2)-7}=\frac{7}{2}$

25. $y=A x^{2}+B x+C$ so $y^{\prime}=2 A x+B$. (explicitly)

$\mathrm{x}=\mathrm{Ay}^{2}+\mathrm{By}+\mathrm{C}$ so (implicitly) $1=2 \mathrm{Ay} \mathbf{y}^{\prime}+\mathrm{B} \mathbf{y}^{\prime}$ and $\mathbf{y}^{\prime}=\frac{1}{2 \mathrm{Ay}+\mathrm{B}}$.

27. $A x^{2}+B x y+C y^{2}+D x+E y+F=0$

      Then $2 A x+B x y^{\prime}+B y+2 C y \mathbf{y}^{\prime}+D+E y^{\prime}=0$ so

      $\mathrm{y}^{\prime}$ $=\frac{\mathrm{-2 A x-B y-D}}{\mathrm{B x+2 C y+E}}$.

29. From problem $23, y^{\prime}=\frac{-2 x-y-3}{x+2 y-7} \quad$ so $y^{\prime}=0$ when $-2 x-y-3=0$ and $y=-2 x-3$.

Substituting $\mathrm{y} =-2 \mathrm{x}-3$ into the original equation, we have

$x^{2}+x(-2 x-3)+(-2 x-3)^{2}+3 x-7(-2 x-3)+4=0$ so

$3 x^{2}+26 x+34=0$ and $x=\frac{-26 \pm \sqrt{26^{2}-4(3)(34)}}{2(3)}=\frac{-26 \pm \sqrt{268}}{6} \approx-1.605$ and $-7.062$.

If $x \approx-1.605$ (point A), then $y=-2 x-3 \approx-2(-1.605)-3=0.21$. Point $A$ is $(-1.605,0.21)$.

If $x \approx-7.062$ (point C), then $y=-2 x-3 \approx-2(-7.062)-3=11.124$. Point $C$ is $(-7.062,11.124)$.

31. From the solution to problem 29, point $\mathrm{C}$ is $(-7.062,11.124)$.

      At $\mathrm{D}, \mathrm{y}^{\prime}=\frac{-2 \mathrm{x}-\mathrm{y}-3}{\mathrm{x}+2 \mathrm{y}-7}$ is undefined so $\mathrm{x}+2 \mathrm{y}-7=0$ and $\mathrm{x}=7-2 \mathrm{y}$.

Substituting $x=7-2 y$ into the original equation, we have

      $(7-2 y)^{2}+(7-2 y) y+y^{2}+3(7-2 y)-7 y+4=0$ so (simplifying) $3 y^{2}-34 y+74=0.$ Then 

      $\mathrm{y}=\frac{34 \pm \sqrt{34^{2}-4(3)(74)}}{2(3)}=\frac{34 \pm \sqrt{268}}{6} \approx 8.398$ and $2.938$.

If $y \approx 2.938$ (point D), then $x=7-2 y \approx 7-2(2.938)=1.124$. Point $D$ is $(1.124,2.938)$. Point B is $(-9.79,8.395)$.

33. (a) $y=\left(x^{2}+5\right)^{7}\left(x^{3}-1\right)^{4}$

     $y^{\prime}=\left(x^{2}+5\right)^{7}(4)\left(x^{3}-1\right)^{3}\left(3 x^{2}\right)+(7)\left(x^{2}+5\right)^{6}(2 x)\left(x^{3}-1\right)^{4}$

     $=\left(x^{2}+5\right)^{6}\left(x^{3}-1\right)^{3}(2 x)\left\{6 x^{3}+30 x+7 x^{3}-7\right\}=\left(x^{2}+5\right)^{6}\left(x^{3}-1\right)^{3}(2 x)\left\{13 x^{3}+30 x-7\right\}$.

(b) $\ln (y)=7 \ln \left(x^{2}+5\right)+4 \ln \left(x^{3}-1\right)$ $\frac{y^{\prime}}{y}=\frac{14 x}{x^{2}+5}+\frac{12 x^{2}}{x^{3}-1}$ so

      $y^{\prime}=y\left\{\frac{14 x}{x^{2}+5}+\frac{12 x^{2}}{x^{3}-1}\right\}=\left(x^{2}+5\right)^{7}\left(x^{3}-1\right)^{4}\left\{\frac{14 x}{x^{2}+5}+\frac{12 x^{2}}{x^{3}-1}\right\}$ and this is the same as in part (a). (Really it is).

35. (a) $y=x^{5}(3 x+2)^{4} \cdot y^{\prime}=x^{5} D\left((3 x+2)^{4}\right)+(3 x+2)^{4} D\left(x^{5}\right)=x^{5} \cdot 4(3 x+2)^{3}(3)+(3 x+2)^{4} \cdot\left(5 x^{4}\right)$

      (b) $\ln (\mathrm{y})=5 \ln (\mathrm{x})+4 \ln (3 \mathrm{x}+2)$. $\frac{y^{\prime}}{y}=\frac{5}{x}+\frac{12}{3 x+2} \quad$ so $y^{\prime}=y\left\{\frac{5}{x}+\frac{12}{3 x+2}\right\}=\left\{x^{5}(3 x+2)^{4}\right\}\left\{\frac{5}{x}+\frac{12}{3 x+2}\right\}$ and this is the same as in part (a).

37. (a) $y=e^{\sin (x)}$ so $y^{\prime}=e^{\sin (x)} \cos (x)$.

      (b) $\ln (y)=\sin (x)$ so $\frac{y^{\prime}}{y}=\cos (x)$ and $y^{\prime}=y \cos (x)=e^{\sin (x)} \cos (x)$.

39. (a) $y=\sqrt{25-x^{2}}$ so $y^{\prime}=\frac{-x}{\sqrt{25-x^{2}}}$.

      (b) $\ln (\mathrm{y})=\frac{1}{2} \ln \left(25-\mathrm{x}^{2}\right)$ so $\frac{\mathrm{y}^{\prime}}{\mathrm{y}}=\frac{1}{2} \frac{-2 \mathrm{x}}{25-\mathrm{x}^{2}}=\frac{-\mathrm{x}}{25-\mathrm{x}^{2}}.$ Then $\mathrm{y}^{\prime}=\mathrm{y} \frac{-\mathrm{X}}{25-\mathrm{x}^{2}}=\sqrt{25-\mathrm{x}^{2}} \frac{-\mathrm{x}}{25-\mathrm{x}^{2}}=\frac{-\mathrm{x}}{\sqrt{25-\mathrm{x}^{2}}}$

41. $y=x^{\cos (x)}$ so $\ln (y)=\cos (x) \cdot \ln (x)$ and $\frac{y^{\prime}}{y}=\cos (x) \cdot \frac{1}{x}-\ln (x) \cdot \sin (x)$

      Then $y^{\prime}=y\left\{\cos (x) \cdot \frac{1}{x}-\ln (x) \cdot \sin (x)\right\}=x^{\cos (x)}\left\{\cos (x) \cdot \frac{1}{x}-\ln (x) \cdot \sin (x)\right\}$

43. $\mathrm{y}=\mathrm{x}^{4}(\mathrm{x}-2)^{7} \sin (3 \mathrm{x})$ so $\ln (\mathrm{y})=4 \ln (\mathrm{x})+7 \ln (\mathrm{x}-2)+\ln (\sin (3 \mathrm{x}))$ and $\frac{\mathrm{y}^{\prime}}{\mathrm{y}}=\frac{4}{\mathrm{x}}+\frac{7}{\mathrm{x}-2}+\frac{3 \cos (3 \mathrm{x})}{\sin (3 \mathrm{x})}$.

      Then $y^{\prime}=y\left\{\frac{4}{x}+\frac{7}{x-2}+\frac{3 \cos (3 x)}{\sin (3 x)}\right\}=x^{4}(x-2)^{7} \sin (3 x)\left\{\frac{4}{x}+\frac{7}{x-2}+\frac{3 \cos (3 x)}{\sin (3 x)}\right\}$.

45. $\ln (y)=x \cdot \ln (3+\sin (x))$ so $\frac{y^{\prime}}{y}=\frac{x \cos (x)}{3+\sin (x)}+\ln (3+\sin (x)).$ Then

      $y^{\prime}=(3+\sin (x))^{x}\left\{\frac{x \cos (x)}{3+\sin (x)}+\ln (3+\sin (x))\right\}$

47. $\begin{array}{l|l} x & f^{\prime}(x) \\ \hline 1 & 1(1.2)=1.2 \\ 2 & 9(1.8)=16.2 \\ 3 & 64(2.1)=134.4 \end{array}$

49. $\begin{array}{c|l} \mathrm{x} & \mathrm{f}^{\prime}(\mathrm{x}) \\ \hline 1 & 5(-1)=-5 \\ 2 & 2(0)=0 \\ 3 & 7(2)=14 \end{array}$

51. $\ln (f \cdot g)=\ln (f)+\ln (g)$ so $\frac{D(f \cdot g)}{f \cdot g}=\frac{f^{\prime}}{f}+\frac{g^{\prime}}{g}$. Then $D(f \cdot g)=(f \cdot g)\left\{\frac{f^{\prime}}{f}+\frac{g^{\prime}}{g}\right\}=f^{\prime} \cdot g+g ' \cdot f$

53. $\ln (f \cdot g \cdot h)=\ln (f)+\ln (g)+\ln (h)$ so $\frac{D\left(f \cdot g \cdot h\right)}{f \cdot g \cdot h}=\frac{f^{\prime}}{f}+\frac{g^{\prime}}{g}+\frac{h^{\prime}}{h}$

55. On your own.