Alex Green
Avishek Khan
Kristina Rowlett
Weldon So
Unit 3 Chapter Test
Mark the solution in red
3.1)Superimpose the graph of the derivative
f(x) = x2+4 @ x=1
Answer:
3.2) The derivative of the following functions fail to be differentiable. Tell why each function is
fails.
a.𝑦 = {
𝑡𝑎𝑛 −1 (𝑥), 𝑥≠0
1,
𝑥=0
}
Discontinuity
b. 𝑦 = 3𝑥 − 2|𝑥| − 1
Corner
5
c.𝑦 = √2 − 𝑥
Vertical Tangent
d. 𝑓(𝑥) = 𝑥 2/3
Cusp
3.3) Find the horizontal tangents of the curve
y=x4-4x2+1
y’=4x3-(4⋅2x)+0
y’=4x3-8x
y’=4(x3⋅2x)
y’=4x(x2-2)
y’=0, ±√2
Find dy/dx
(𝑥+1)(𝑥+2)
y=
(𝑥−1)(𝑥−2)
Derivative of:
x+1 is 1
x+2 is 1
x-1 is 1 x-2 is 1
By the Product Rule:
(1)(𝑥+1)+(1)(𝑥+2)
(1)(𝑥−1)+(1)(𝑥−2)
2𝑥+3
2𝑥−3
Derivative of:
2x+3 is 2
2x-3 is 2
By the Quotient Rule:
(2)(2𝑥−3)−(2)(2𝑥+3)
(2𝑥−3)2
4𝑥−6−4𝑥−6
(2𝑥−3)2
−12
y’=
(2𝑥−3)2
3.4)Velocity and Other Rates of Change
Solve the following parts of the problem using what you know about rates of change:
A. Write the area A of a square as a function of the side length s.
B. Find the (instantaneous) rate of change of the area A with respect to a side s.
C. Evaluate the rate of change of A at s = 1 and s = 5.
D. If s is measured in centimeters and A is measured in square centimeters, what units
would be appropriate for
𝑑𝐴
?
𝑑𝑠
A) Since the area of a square with respect to a side length is 𝑠 2 , then the area as a
function with respect side s is 𝐴 = 𝑠 2
B) The instantaneous rate of change is the same as the first derivative. Thus, it is 𝐴′ = 2𝑠
C) 𝐴′ (1) = 2(1) = 2 ; 𝐴′ (5) = 2(5) = 10
D)
𝑐𝑚2
𝑐𝑚
3.5)Find the equation of the lines tangent and normal to the graph of y= sin x + 3 at x = π
Answer:
3.6)Differentiate 𝑦 = 3 𝑡𝑎𝑛√𝑥.
y’ = 3(sec2 √𝑥) t(√𝑥)
y’ = 3(sec2 √𝑥) t(x1/2)
y’ = 3(sec2 √𝑥) (½)(x-1/2)
1
)
𝑥 1/2
y’ = 3(sec2 √𝑥) (½)(
y’ =
3(𝑠𝑒𝑐2 √𝑥)
2√𝑥
3.7)Find dy/dx and find the slope of the curve at the indicated point
x2+y2=13 @ (-2, 3)
2x+2y(dy/dx)=0
2x=-2y(dy/dx)
-2x/2y=(dy/dx)
-x/y=(dy/dx)
-(-2)/(3)=(dy/dx)
Slope=⅔
Find the lines that are (a) tangent and (b) normal to the curve at the given point
x2+xy-y2=1 @ (2, 3)
Derivative of xy:
Using the product rule
𝑑𝑦
𝑑𝑥
( )(x)+(1)(y)
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑦
2x+y=- x+ 2y
𝑑𝑥
𝑑𝑥
𝑑𝑦
2x+y=( )(-x+2y)
𝑑𝑥
𝑑𝑦 2𝑥+𝑦
=
𝑑𝑥 −𝑥+2𝑦
2x+ x+y-( )2y=0
Slope of Tangent Line:
2(2)+(3) 7
=
−(2)+2(3) 4
Slope of Normal Line:
-
4
7
Tangent Line:
7
4
y-3=( )(x-2)
Normal Line:
4
y-3=(-7)(x-2)
3.8) Find the derivative of 𝑦 = 𝑐𝑜𝑠 −1 (8𝑥 4 )with respect to x
The derivative of 𝑜𝑠 −1 (𝑥) =
𝑐𝑜𝑠 −1 (8𝑥 4 ) =
−1
√1−(8𝑥 4 )2
√1−𝑥 2
3
x 32𝑥
3.9) FInd the derivative of
Answer:
−1
. Thus, the derivative with respect to x is:
Chapter 1 Review)Solve for x
1 + 𝑠𝑖𝑛(𝑥) = 2𝑐𝑜𝑠
1+sinx = 2(1-sin2 x)
1+sinx = 2-2sin2 x
2sin2 -sinx -1=0
(2sinx-1)(sinx+1)=0
2sinx-1 =0
or
sinx=-½
𝜋 5𝜋
6
x= 6 ,
or
x=
3𝜋
2
sinx+1=0
sinx=-1
2
(𝑥)
Vectors)
Solution on Mr. Wahbeh’s Website
Solutions on Mr. Wahbeh’s website
Chapter 2)
Find the limit of:
When you look at the graph of the function, there is a vertical asymptote at x=5. As the line
approaches 5 from the left, the value heads towards negative infinity. Hence, the limit is
negative infinity.