# 3.5E: Trig Derivatives Exercises

## 3.5: Derivatives of Trigonometric Functions

### Exercise:

For the following exercises, find (frac{dy}{dx}) for the given functions.

175) (y=x^2−secx+1)

(frac{dy}{dx}=2x−secxtanx)

176) (y=3cscx+frac{5}{x})

177) (y=x^2cotx)

(frac{dy}{dx}=2xcotx−x^2csc^2x)

178) (y=x−x^3sinx)

179) (y=frac{secx}{x})

(frac{dy}{dx}=frac{xsecxtanx−secx}{x^2})

180) (y=sinxtanx)

181) (y=(x+cosx)(1−sinx))

(frac{dy}{dx}=(1−sinx)(1−sinx)−cosx(x+cosx))

182) (y=frac{tanx}{1−secx})

183) (y=frac{1−cotx}{1+cotx})

(frac{dy}{dx}=frac{2csc^2x}{(1+cotx)^2})

184) (y=cosx(1+cscx))

For the following exercises, find the equation of the tangent line to each of the given functions at the indicated values of (x). Then use a calculator to graph both the function and the tangent line to ensure the equation for the tangent line is correct.

185) ([T] f(x)=−sin{x},x=0)

(y=−x)

186) ([T] f(x)=cscx,x=frac{π}{2})

187) ([T] f(x)=1+cosx,x=frac{3π}{2})

(y=x+frac{2−3π}{2})

188) ([T] f(x)=secx,x=frac{π}{4})

189) ([T] f(x)=x^2− an{x}=0)

(y=−x)

190) ([T] f(x)=5cotxx=frac{π}{4})

For the following exercises, find (frac{d^2y}{dx^2}) for the given functions.

191) (y=xsinx−cosx)

(3cosx−xsinx)

192) (y=sinxcosx)

193) (y=x−frac{1}{2}sinx)

(frac{1}{2}sinx)

194) (y=frac{1}{x}+tanx)

195) (y=2cscx)

(csc(x)(3csc^2(x)−1+cot^2(x)))

196) (y=sec^2x)

197) Find all (x) values on the graph of (f(x)=−3sinxcosx) where the tangent line is horizontal.

(frac{(2n+1)π}{4}),where (n) is an integer

198) Find all (x) values on the graph of (f(x)=x−2cosx) for (0

199) Let (f(x)=cotx.) Determine the points on the graph of (f) for (0

((frac{π}{4},1),(frac{3π}{4},−1))

200) [T] A mass on a spring bounces up and down in simple harmonic motion, modeled by the function (s(t)=−6cost) where s is measured in inches and t is measured in seconds. Find the rate at which the spring is oscillating at (t=5) s.

201) Let the position of a swinging pendulum in simple harmonic motion be given by (s(t)=acost+bsint). Find the constants (a) and (b) such that when the velocity is 3 cm/s, (s=0) and (t=0).

(a=0,b=3)

202) After a diver jumps off a diving board, the edge of the board oscillates with position given by (s(t)=−5cost) cm at (t)seconds after the jump.

a. Sketch one period of the position function for (t≥0).

b. Find the velocity function.

c. Sketch one period of the velocity function for (t≥0).

d. Determine the times when the velocity is 0 over one period.

e. Find the acceleration function.

f. Sketch one period of the acceleration function for (t≥0).

203) The number of hamburgers sold at a fast-food restaurant in Pasadena, California, is given by (y=10+5sinx) where (y) is the number of hamburgers sold and x represents the number of hours after the restaurant opened at 11 a.m. until 11 p.m., when the store closes. Find (y') and determine the intervals where the number of burgers being sold is increasing.

(y′=5cos(x)), increasing on ((0,frac{π}{2}),(frac{3π}{2},frac{5π}{2})), and ((frac{7π}{2},12))

204) [T] The amount of rainfall per month in Phoenix, Arizona, can be approximated by (y(t)=0.5+0.3cost), where t is months since January. Find (y′)and use a calculator to determine the intervals where the amount of rain falling is decreasing.

For the following exercises, use the quotient rule to derive the given equations.

205) (frac{d}{dx}(cotx)=−csc^2x)

206) (frac{d}{dx}(secx)=secxtanx)

207) (frac{d}{dx}(cscx)=−cscxcotx)

208) Use the definition of derivative and the identity (cos(x+h)=cosxcosh−sinxsinh)

to prove that (frac{d(cosx)}{dx}=−sinx).

For the following exercises, find the requested higher-order derivative for the given functions.

209) (frac{d^3y}{dx^3}) of (y=3cosx)

(3sinx)

210) (frac{d^2y}{dx^2}) of (y=3sinx+x^2cosx)

211) (frac{d^4y}{dx^4}) of (y=5cosx)

(5cosx)

212) (frac{d^2y}{dx^2}) of (y=secx+cotx)

213) (frac{d^3y}{dx^3}) of (y=x^{10}−secx)

(720x^7−5tan(x)sec^3(x)−tan^3(x)sec(x))

## 3.5E: Trig Derivatives Exercises

(Recall that . The product rule is NOT necessary here.)

SOLUTION 2 : Differentiate . Apply the product rule.

SOLUTION 3 : Differentiate . Apply the quotient rule.

(Recall the well-known trigonometry identity .)

SOLUTION 4 : Differentiate . Apply the product rule.

SOLUTION 5 : Differentiate . To avoid using the chain rule, first rewrite the problem as

Now apply the product rule. Then

SOLUTION 6 : Differentiate . To avoid using the chain rule, recall the trigonometry identity , and first rewrite the problem as

Now apply the product rule twice. Then

(This is an acceptable answer. However, an alternative answer can be gotten by using the trigonometry identity .)

SOLUTION 7 : Differentiate . Rewrite g as a triple product and apply the triple product rule. Then

so that the derivative is

SOLUTION 8 : Evaluate . It may not be obvious, but this problem can be viewed as a differentiation problem. Recall that

If , then , and letting it follows that

SOLUTION 9 : Differentiate . Apply the chain rule to both functions. (If necessary, review the section on the chain rule .) Then

SOLUTION 10 : Differentiate . This is NOT a product of functions. It's a composition of functions. Apply the chain rule. Then

SOLUTION 11 : Differentiate . Apply the quotient rule first, followed by the chain rule. Then

SOLUTION 12 : Differentiate . Apply the product rule first, followed by the chain rule. Then

SOLUTION 13 : Differentiate . Apply the chain rule four times ! Then

SOLUTION 14 : Differentiate . Apply the quotient rule first. Then

(Apply the product rule in the first part of the numerator.)

SOLUTION 15 : Find an equation of the line tangent to the graph of at x =-1 . If x = -1 then so that the tangent line passes through the point (-1, 0 ) . The slope of the tangent line follows from the derivative

The slope of the line tangent to the graph at x = -1 is

Thus, an equation of the tangent line is

y - 0 = -2 ( x - (-1) ) or y = -2 x - 2 .

SOLUTION 16 : Find an equation of the line perpendicular to the graph of at . If then so that the tangent line passes through the point . The slope of the tangent line follows from the derivative of y . Then

The slope of the line tangent to the graph at is

Thus, the slope of the line perpendicular to the graph at is

so that an equation of the line perpendicular to the graph at is

SOLUTION 17 : Assume that . Solve f '( x ) = 0 for x in the interval . Use the chain rule to find the derivative of f . Then

## Derivatives of Inverse Trigonometric Functions

In the previous topic, we have learned the derivatives of six basic trigonometric functions:

In this section, we are going to look at the derivatives of the inverse trigonometric functions , which are respectively denoted as

The inverse functions exist when appropriate restrictions are placed on the domain of the original functions.

For example, the domain for (arcsin x) is from (-1) to (1.) The range, or output for (arcsin x) is all angles from ( – large<2>> ormalsize) to (large<2>> ormalsize) radians.

The domains of the other trigonometric functions are restricted appropriately, so that they become one-to-one functions and their inverse can be determined.

### Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions can be obtained using the inverse function theorem. For example, the sine function (x = varphi left( y ight) ) (= sin y) is the inverse function for (y = fleft( x ight) ) (= arcsin x.) Then the derivative of (y = arcsin x) is given by

Using this technique, we can find the derivatives of the other inverse trigonometric functions:

In the last formula, the absolute value (left| x ight|) in the denominator appears due to the fact that the product (< an ysec y>) should always be positive in the range of admissible values of (y), where (y in left( <0,<2> ormalsize>> ight) cup left( <<2> ormalsize>,pi > ight),) that is the derivative of the inverse secant is always positive.

Similarly, we can obtain an expression for the derivative of the inverse cosecant function:

### Table of Derivatives of Inverse Trigonometric Functions

The derivatives of (6) inverse trigonometric functions considered above are consolidated in the following table:

The equation of a circular motion is: φ(t) = ½t². What is the angular velocity and the acceleration at the seven second mark?

A man is 2000 m from the base of a tower and is launching a rocket in the direction of the same tower. When the rocket takes off the change in the angle between the flight path and the land is represented by Φ(t) according to time. Knowing that Φ'(t) = Π/3, determine:

1. The height of the rocket when Φ = Π/3 radians.

2. The velocity of the rocket when Φ = Π / 3 radians?

## Exercise 4

Differentiate the following exponential functions:

1)

2)

3)

4)

5)

• Given an implicit function with the dependent variable y and the independent variable x (or the other way around).
• Differentiate the entire equation with respect to the independent variable (it could be x or y).
• After differentiating, we need to apply the chain rule of differentiation.
• Solve the resultant equation for dy/dx (or dx/dy likewise) or differentiate again if the higher-order derivatives are needed.

“Some function of y and x equals to something else”. Knowing x does not help us compute y directly. For instance,

x 2 + y 2 = r 2 ( Implicit function)

Differentiate with respect to x:
d(x 2 ) /dx + d(y 2 )/ dx = d(r 2 ) / dx

Solve each term:

Using Power Rule: d(x 2 ) / dx = 2x

Using Chain Rule : d(y 2 )/ dx = 2y dydx

r 2 is a constant, so its derivative is 0: d(r 2 )/ dx = 0

Which gives us:

2x + 2y dy/dx = 0

Collect all the dy/dx on one side

y dy/dx = −x

Solve for dy/dx:

dy/dx = −xy

## Trigonometry

We have seen that the sine and cosine functions can be constructed geometrically in terms of a unit circle centered at the origin. This applet shows the relationship between the values of the sine, cosine and tangent on the unit circle and their respective graphs. Once the applet is open, click on the exercises button for a full explanation.

Can't see the above java applet? Click here to see how to enable Java on your web browser. (This applet was developed by Franz Embacher and Petra Oberhuemer (maths online) and is used here with permission)

The graphs of sin x, cos x and tan x are periodic. A periodic function is one that repeats its values after a period has been added to the independent variable, in this case x. The functions sin x and cos x both have periods equal to 2&pi. That is, sin x = sin(x + 2&pi) = sin(x + 4&pi) = sin(x + 2k&pi) for any integer k and cosx = cos(x + 2&pi) = cos(x + 4&pi) = cos(x + 2k&pi) for any integer k. Their graphs are each shown below. Notice that the distance between successive peaks (or successive troughs) in the graphs of sin x and cos x is equal to the period 2&pi. This distance is known in physics as the wavelength. The amplitude of each of the functions sin x and cos x is 1. This refers to the distance from the peak (or trough) and the baseline (the horizontal line located halfway between the peak and trough, in this case the x-axis) or half the vertical distance from the peak to trough.

The graph of tan x is periodic but the period is &pi. So tan x = tan(x + k&pi) for any integer k.

For the function y = a*sin(b*x + c) + d, the amplitude is given by the value of the parameter a. The period of the function is given by 2&pi/b. The parameter c shifts the function horizontally to the left by the given amount and d shifts the function vertically upwards by the given amount. Use the applet below to see visually how this works. Move the sliders to adjust the values of a, b, c and d in y = a*sin(b*x+c)+d.

For the function y = a*cos(b*x + c) + d, the parameters a, b, c and d have the same effect on the cosine function as they did on the sine. Use the applet below to see visually how this works. Move the sliders to adjust the values of a, b, c and d in y = a*cos(b*x+c)+d.

### Exercise Applets

Match the functions of the form a sin(b + c x) or a cos(b + c x) with their graphs. The applet is started by clicking on the red button and will open in its own window.

Match the graphs of the functions of the form a sin(b + c x) or a cos(b + c x) with their functions. The applet is started by clicking on the red button and will open in its own window.

## Chapter 3 Class 11 Trigonometric Functions

NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below.

We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn

• What is a positive or a negative angle
• Measuring angles in Degree, Minutes and Seconds
• Radian measure of an angle
• Converting Degree to Radians, and vice-versa
• Sign of sin, cos, tan in all 4 quadrants
• Finding values of trigonometric functions when one value is given (Example: Finding value of sin, cot, cosec, tan, sec, when cos x = -3/5 is given)
• Finding Value of trigonometric functions, given angle
• Solving questions by formula like (x + y) formula, 2x 3x formula, Cos x + cos y formula , 2 sin x sin y formula
• Finding principal and general solutions of a trigonometric equation
• Sin and Cosine Formula with supplementary Questions

Important questions are marked, and Formula sheet is also provided. Click on an exercise or topic to begin.

## 3.5E: Trig Derivatives Exercises

Some of the following problems require use of the chain rule.

## 3.5E: Trig Derivatives Exercises

A Tangent Line is a line which touches a curve at one and only one point.

To find the equation of tangent line at a point (x1, y1),  we use the formula

Here m is slope at (x1, y1) and (x1, y1) is the point at which we draw a tangent line.

find the equation of the tangent line at x  =  5.

We should find equation of tangent at the point (5, 3).

Equation of the tangent line :

Tangent line at the point (5, 3).

Equation of the tangent line at the point (5, 3) is

is tangent to the graph of the function f ਊt  (2, 15),  what is f' (2) ?

Equation of the tangent line :

What is the x -coordinate  of the point where the tangent line to

Since the tangent line drawn for the given curve is parallel to x-axis, slope of the required tangent line is 0.

So, the required x-co ordinate is x  =  -6.

Find the equation of the tangent line which goes through the point (2, -1) and is parallel to the line given by the equation 2x-y  =  1

Since the required tangent line is parallel to the given line 2x-y  =  1, slope of the given line is equal to the slope of the tangent line.

m  =  -coefficient of x/coefficient of y

Equation of the tangent line :

For a particular g, we know g'(5)  =  2 and g(5)  =  3. Write an equation of the tangent to g at x  =  5.

Slope at the point x  =  5 is 2.

We draw the tangent line at the point (5, 3).

Equation of the tangent line :

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