2.7.3: Additional Exercises

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Reading Questions

What is the difference between an explicit function and an implicit function? Give an example of each.
What is the main goal of implicit differentiation?
When using implicit differentiation to find $\frac{dy}{dx}$, what rule must be applied every time a term involving $y$ is differentiated with respect to $x$?
If you differentiate the term $y^3$ with respect to $x$, what is the result?
If you differentiate the term $xy$ with respect to $x$, what rule must you use and what is the result?
After differentiating both sides of an implicitly defined equation with respect to $x$, what is the next general step to find $\frac{dy}{dx}$?
In Example 1, where $x^2+y^2=25$, what was $\frac{dy}{dx}$ found to be?
Is the expression for $\frac{dy}{dx}$ found through implicit differentiation typically in terms of $x$ only, $y$ only, or both $x$ and $y$?
How can implicit differentiation be used to find the equation of a tangent line to a curve at a given point?
What is the "Great Advice to Save Time and Prevent Mistakes" regarding finding the slope at a specific point?
How can implicit differentiation be used to find a second derivative, $\frac{d^2y}{dx^2}$?
What is the derivative of $f(x) = \sin^{-1}x$? Are there any restrictions on $x$?
What is the derivative of $f(x) = \tan^{-1}x$?
The derivatives of inverse trigonometric functions are what type of functions (e.g., trigonometric, algebraic, exponential)?
What critical step involving function ranges must be recalled when proving the derivative of an inverse trigonometric function like $\arcsin x$?
What is the derivative of $f(x) = \text{sinh}^{-1}x$?
What is the derivative of $f(x) = \text{tanh}^{-1}x$?
Which two inverse hyperbolic functions have the same derivative formula, $\frac{1}{1-x^2}$, and how do their domains differ for this derivative?
How is implicit differentiation used to find the derivatives of inverse hyperbolic functions?

Homework

In exercises 1 - 10, use implicit differentiation to find $\dfrac{dy}{dx}$.

1) $x^2−y^2=4$

2) $6x^2+3y^2=12$

Answer: $\dfrac{dy}{dx}=\dfrac{−2x}{y}$

3) $x^2y=y−7$

4) $3x^3+9xy^2=5x^3$

Answer: $\dfrac{dy}{dx}=\dfrac{x}{3y}−\dfrac{y}{2x}$

5) $xy−\cos(xy)=1$

6) $y\sqrt{x+4}=xy+8$

Answer: $\dfrac{dy}{dx}=\dfrac{y−\dfrac{y}{2\sqrt{x+4}}}{\sqrt{x+4}−x}$

7) $−xy−2=\frac{x}{7}$

8) $y\sin(xy)=y^2+2$

Answer: $\dfrac{dy}{dx}=\dfrac{y^2\cos(xy)}{2y−\sin(xy)−xy\cos(xy)}$

9) $(xy)^2+3x=y^2$

10) $x^3y+xy^3=−8$

Answer: $\dfrac{dy}{dx}=\dfrac{−3x^2y−y^3}{x^3+3xy^2}$

In exercises 11 - 26, find the derivatives for the functions.

11) $y=\dfrac{1}{\tan^{−1}(x)}$

Answer: $\dfrac{dy}{dx} = \dfrac{−1}{(1+x^2)(\tan^{−1}x)^2}$

12) $y=x \cdot \csc^{−1}x$

13) $\tanh^{−1}(4x)$

Answer: $\dfrac{4}{1−16x^2}$

14) $\sinh^{−1}(x^2)$

15) $\sinh^{−1}(\cosh(x))$

Answer: $\dfrac{\sinh(x)}{\sqrt{\cosh^2(x)+1}}$

16) $\cosh^{−1}(x^3)$

17) $\tanh^{−1}(\cos(x))$

Answer: $−\csc(x)$

18) $e^{\sinh^{−1}(x)}$

19) $y=\sin^{−1}(x^2)$

Answer: $\dfrac{dy}{dx} = \dfrac{2x}{\sqrt{1−x^4}}$

20) $y=\sec^{−1}(−x)$

21) $y=\cos^{−1}(2x) \cdot \sin^{−1}(2x)$

22) $y=\cos^{−1}\left(\sqrt{x}\right)$

23) $y=\sec^{−1}\left(\frac{1}{x}\right)$

Answer: $\dfrac{dy}{dx} = \dfrac{−1}{\sqrt{1−x^2}}$

24) $y=\sqrt{\csc^{−1}x}$

25) $y=(1+\tan^{−1}x)^3$

Answer: $\dfrac{dy}{dx} = \dfrac{3(1+\tan^{−1}x)^2}{1+x^2}$

26) $y=\cot^{−1}\sqrt{4−x^2}$

Answer: $\dfrac{dy}{dx} = \dfrac{x}{(5−x^2)\sqrt{4−x^2}}$

For exercises 27 - 32, find the equation of the tangent line to the graph of the given equation at the indicated point. Use a calculator or computer software to graph the function and the tangent line.

27) [Technology Required] $x^4y−xy^3=−2, \quad (−1,−1)$

28) [Technology Required] $x^2y^2+5xy=14,\quad (2,1)$

Answer

$y=−\frac{1}{2}x+2$

$The graph has a crescent in each of the four quadrants. There is a straight line marked T(x) with slope −1/2 and y intercept 2.$

29) [Technology Required] $\tan(xy)=y,\quad \left(\frac{ \pi }{4},1\right)$

30) [Technology Required] $xy^2+\sin( \pi y)−2x^2=10, \quad (2,−3)$

Answer

$y=\frac{1}{ \pi +12}x−\frac{3 \pi +38}{ \pi +12}$

$The graph has two curves, one in the first quadrant and one in the fourth quadrant. They are symmetric about the x axis. The curve in the first quadrant goes from (0.3, 5) to (1.5, 3.5) to (5, 4). There is a straight line marked T(x) with slope 1/( \pi + 12) and y intercept −(3 \pi + 38)/( \pi + 12).$

31) [Technology Required] $\dfrac{x}{y}+5x−7=−\frac{3}{4}y, \quad (1,2)$

32) [Technology Required] $xy+\sin(x)=1,\quad \left(\frac{ \pi }{2},0\right)$

Answer

$y=0$

$The graph starts in the third quadrant near (−5, 0), remains near 0 until x = −4, at which point it decreases until it reaches near (0, −5). There is an asymptote at x = 0. The graph begins again near (0, 5) decreases to (1, 0) and then increases a little bit before decreasing to be near (5, 0). There is a straight line marked T(x) that coincides with y = 0.$

33) Find the equation of the tangent line to the graph of the equation $\sin^{−1}x+\sin^{−1}y=\frac{ \pi }{6}$ at the point $\left(0,\frac{1}{2}\right)$.

34) Find the equation of the tangent line to the graph of the equation $\tan^{−1}(x+y)=x^2+\frac{ \pi }{4}$ at the point $(0,1)$.

Answer: $y=−x+1$

35) [Technology Required] The graph of a folium of Descartes with equation $2x^3+2y^3−9xy=0$ is given in the following graph.

$A folium is graphed which has equation 2x3 + 2y3 – 9xy = 0. It crosses over itself at (0, 0).$

a. Find the equation of the tangent line at the point $(2,1)$. Graph the tangent line along with the folium.

b. Find the equation of the normal line to the tangent line in a. at the point $(2,1)$.

36) For the equation $x^2+2xy−3y^2=0$,

a. Find the equation of the normal to the tangent line at the point $(1,1)$.

b. At what other point does the normal line in a. intersect the graph of the equation?

Answer: a. $y=−x+2$
b. $(3,−1)$

37) Find all points on the graph of $y^3−27y=x^2−90$ at which the tangent line is vertical.

38) For the equation $x^2+xy+y^2=7$,

a. Find the $x$-intercept(s).

b.Find the slope of the tangent line(s) at the $x$-intercept(s).

c. What does the value(s) in part b. indicate about the tangent line(s)?

Answer: a. $\left( \pm \sqrt{7},0\right)$
b. $−2$
c. They are parallel since the slope is the same at both intercepts.

39) Find $y^{\prime}$ and $y^{\prime\prime}$ for $x^2+6xy−2y^2=3$.

40) [Technology Required] The number of cell phones produced when $x$ dollars is spent on labor and $y$ dollars is spent on capital invested by a manufacturer can be modeled by the equation $60x^{3/4}y^{1/4}=3240$.

a. Find $\frac{dy}{dx}$ and evaluate at the point $(81,16)$.

b. Interpret the result of a.

Answer: a. $\frac{dy}{dx}=−0.5926$
b. When $81 is spent on labor and $16 is spent on capital, the amount spent on capital is decreasing by $0.5926 per $1 spent on labor.

41) [Technology Required] The number of cars produced when $x$ dollars is spent on labor and $y$ dollars is spent on capital invested by a manufacturer can be modeled by the equation $30x^{1/3}y^{2/3}=360$.

(Both $x$and $y$ are measured in thousands of dollars.)

a. Find $\frac{dy}{dx}$ and evaluate at the point $(27,8)$.

b. Interpret the result of part a.

42) The volume of a right circular cone of radius $x$ and height $y$ is given by $V=\frac{1}{3} \pi x^2y$. Suppose that the volume of the cone is $85 \pi \,\text{cm}^3$. Find $\dfrac{dy}{dx}$ when $x=4$ and $y=16$.

Answer: $\dfrac{dy}{dx} = −8$

For exercises 43 - 44, consider a closed rectangular box with a square base with side $x$ and height $y$.

43) Find an equation for the surface area of the rectangular box, $S(x,y)$.

44) If the surface area of the rectangular box is 78 square feet, find $\dfrac{dy}{dx}$ when $x=3$ feet and $y=5$ feet.

Answer: $\dfrac{dy}{dx} = −2.67$

In exercises 45 - 47, use implicit differentiation to determine $y′$.

45) $x=\sin y$

46) $x=\cos y$

Answer: $y′=−\dfrac{1}{\sqrt{1−x^2}}$

47) $x=\tan y$

48) [Technology Required] The position of a moving hockey puck after $t$ seconds is $s(t) = \tan^{−1}t$ where $s$ is in meters.

a. Find the velocity of the hockey puck at any time $t$.

b. Find the acceleration of the puck at any time $t$.

c. Evaluate parts a. and b. for $t=2,\, 4$,and $6$ seconds.

d. What conclusion can be drawn from the results in c.?

Answer: a. $v(t)=\dfrac{1}{1+t^2}$
b. $a(t)=\dfrac{−2t}{(1+t^2)^2}$
c. (a) $0.2,\, 0.06,\, 0.03$; (b) $−0.16,\, −0.028,\, −0.0088$
d. The hockey puck is decelerating/slowing down at 2, 4, and 6 seconds.

49) [Technology Required] A building that is 225 feet tall casts a shadow of various lengths $x$ as the day goes by. An angle of elevation $ \theta $ is formed by lines from the top and bottom of the building to the tip of the shadow, as seen in the following figure. Find the rate of change of the angle of elevation $\frac{d \theta }{dx}$ when $x=272$ feet.

$A building is shown with height 225 ft. A triangle is made with the building height as the opposite side from the angle \theta . The adjacent side has length x.$

50) [Technology Required] A pole stands 75 feet tall. An angle $ \theta $ is formed when wires of various lengths of $x$ feet are attached from the ground to the top of the pole, as shown in the following figure. Find the rate of change of the angle $\frac{d \theta }{dx}$ when a wire of length 90 feet is attached.

$A flagpole is shown with height 75 ft. A triangle is made with the flagpole height as the opposite side from the angle \theta . The hypotenuse has length x.$

Answer: $−0.0168$ radians per foot

51) [Technology Required] A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by $ \theta =\tan^{−1}\left(\frac{x}{2000}\right)$, where $x$ is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5000 feet apart.

$A rocket is shown with in the air with the distance from its nose to the ground being x. A triangle is made with the rocket height as the opposite side from the angle \theta . The adjacent side has length 2000.$

52) [Technology Required] A local movie theater with a 30-foot-high screen that is 10 feet above a person’s eye level when seated has a viewing angle $ \theta $ (in radians) given by $ \theta =\cot^{−1}\frac{x}{40}−\cot^{−1}\frac{x}{10}$, where $x$ is the distance in feet away from the movie screen that the person is sitting, as shown in the following figure.

$A person is shown with a right triangle coming from their eye (the right angle being on the opposite side from the eye), with height 10 and base x. There is a line drawn from the eye to the top of the screen, which makes an angle \theta with the triangle’s hypotenuse. The screen has a height of 30.$

a. Find $\dfrac{d \theta }{dx}$.

b. Evaluate $\dfrac{d \theta }{dx}$ for $x=5,\,10,\,15$, and $20$.

c. Interpret the results in part b.

d. Evaluate $\dfrac{d \theta }{dx}$ for $x=25,\,30,\,35$, and $40$.

e. Interpret the results in part d. At what distance $x$ should the person stand to maximize his or her viewing angle?

Answer: a. $\dfrac{d \theta }{dx}=\dfrac{10}{100+x^2}−\dfrac{40}{1600+x^2}$
b. $\frac{18}{325},\,\frac{9}{340},\,\frac{42}{4745},\,0$
c. As a person moves farther away from the screen, the viewing angle is increasing, which implies that as he or she moves farther away, his or her screen vision is widening. d. $−\frac{54}{12905},\,−\frac{3}{500},\,−\frac{198}{29945},\,−\frac{9}{1360}$
e. As the person moves beyond 20 feet from the screen, the viewing angle is decreasing. The optimal distance the person should stand for maximizing the viewing angle is 20 feet.

53) Prove the formula for the derivative of $y=\sinh^{−1}(x)$ by differentiating $x=\sinh(y)$.

54) Prove the formula for the derivative of $y=\cosh^{−1}(x)$ by differentiating $x=\cosh(y)$.

55) Prove the formula for the derivative of $y=\text{sech}^{−1}(x)$ by differentiating $x=\text{sech}(y)$.

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