Math Practice Test- Unit 3
By: Zachary Meicho, Selma Ducanovic, Marc Campagnaro, and Rachel Larson
3.1 Use the definition
indicated point.
a)
to find the derivative of the given function at the
, a = -1
b)
3.2 a) Find all values of x for which the function is differentiable.
b) Tell whether the problem is a corner, a cusp, a vertical tangent, or a discontinuity.
3.3 Find the derivative.
a)
b)
3.4
1.
a)
b)
c)
d)
When does the body reverse direction?
When (approximately) is the body moving at a constant rate?
Graph the body’s speed for 0 ≤ t ≤ 10
Graph the acceleration, where defined.
2.
The position of a body at time t sec is S = t3 – 6t2 + 9t m. Find the body’s acceleration
each time the velocity is zero.
3.5
1.
2.
3.6
1.
2.
The position function of a weight hanging from a spring is s= 5 sinx. Find the velocity and
acceleration at time t.
Find equations for the lines that are tangent and normal to the curve y=√2 cosx at the
point (pi/4, 1)
Using the Chain rule and the product rule find the derivative of 𝑠𝑖𝑛3 (𝑥) 𝑡𝑎𝑛(4𝑥)
Use the chain rule to solve for y= cos(sin x); u= sin x
3.7
1.
(a) Find the derivative of 𝑦 2 = 𝑥 2 + 2x
(b) Find the double derivative
2.
Using implicit differentiation find the derivative of 𝑥 2 + 𝑦 2 = 35
3.8
1.
Find the derivative of y=sec-1(2s3+4) with respect to s.
2.
(a) Find an equation for the line tangent to the graph of y= 2 tan(4x) at the point
(𝜋/16,2).
(a) Find an equation for the line tangent to the graph of y= ¼ tan-1(½x) at the point
(2,𝜋/16).
3.9
1.
2.
f(x)= 9ex
Find the derivative of y with respect to x of y= ln (x5).
Worked out Solutions:
3.1 a)
First, apply the definition to the given function.
Simplify.
Substitute 0 for h.
b)
First, apply the definition to the given function.
Simplify and substitute 0 for h.
3.2 a)
Expand the fractions, and you will see that there are discontinuities at x = -1 and x = 5. A function is not
differentiable where there is a discontinuity.
b) Corner. The function forms two different lines to the left and right of x = 0. Where these lines
meet they form a corner.
3.3 a)
b)
Product rule
Quotient rule
3.4
1.
*See Graph*
a) Body reverses direction at t = 2 and t = 7
b) At a constant speed between t = 3 and t = 6
c)
d)
2.
2
S’= V = 3t – 12t + 9
2
2
0 = 3t – 12t + 9 = 3(t – 4t + 3)
0 = (t – 1)(1 – 3)
V(0) = 1, 3
S’’= a = 6t – 12
a(1) = -6 m/sec2
a(3) = 6 m/sec2
3.5
1.
2.
S=5sinx
*Use derivative of sin/cos rule*
S’=5cosx
S’’= -5sinx
y=√2cosx
y’= -√2sinx
y’ = -√2sin(pi/4) = -√2(√2/2) = -1
Tangent: y - 1 = -1(x - (pi/4))
Normal: y - 1 = 1 (x - (pi/4))
3.6
1.
𝑦 = 𝑠𝑖𝑛3 (𝑥) 𝑡𝑎𝑛(4𝑥)
y’= [(3𝑠𝑖𝑛2 (𝑥)) ⋅ 𝑐𝑜𝑠(𝑥)] (𝑡𝑎𝑛(4𝑥))+ (𝑠𝑖𝑛3 (𝑥)) (𝑠𝑒𝑐 2 (4𝑥)) ⋅(4)
y’= 3𝑠𝑖𝑛2 (𝑥) 𝑐𝑜𝑠(𝑥) 𝑡𝑎𝑛(4𝑥) + 4𝑠𝑖𝑛3 (𝑥)𝑠𝑒𝑐 2 (4𝑥)
2.
y=cos(sin x)
y= -sin(sin x) ⋅cos x
y= -sin(sin x) cos(x)
3.7
1.
(a) 𝑦 2 = 𝑥 2 + 2x
2yy’=
y’=
2𝑦
𝑥+1
𝑦
𝑥+1
𝑦
(b) y’’=
2.
2𝑥 + 2
y’’ =
(1) (𝑦) − (𝑥 + 1) (𝑦′)
𝑦2
y’’ =
𝑦− [ (𝑥 + 1) (𝑥 + 1 /𝑦)]
𝑦 2
y’’ =
𝑦− (𝑥+ 1) 2 /𝑦
𝑦 2
y’’ =
𝑦 2 − (𝑥 + 1)2
𝑦3
y’’ =
(𝑥 2 + 2𝑥) − (𝑥 2 + 2𝑥 + 1)
𝑦3
y’’ =
−1
𝑦3
𝑥 2 + 𝑦 2 = 35
2x + 2yy’= 0
2yy’= -2x
y’=
−2𝑥
2𝑦
y’=
−𝑥
𝑦
3.8
1.
Find the derivative of y=sec-1(2s3+4) with respect to s.
yI=
yI=
2.
1𝑥 6𝑠2
|2𝑠3 +4
|√(2𝑠3 +4)2 −1
6𝑠2
|2𝑠3 +4
|√4𝑠6 +16𝑠3 +15
(a) Find an equation for the line tangent to the graph of y= 2 tan(4x) at the point
(𝜋/16,2).
(a) Find an equation for the line tangent to the graph of y= ¼ tan-1(½x) at the point
(2,𝜋/16).
A tangent line to a curve at a given point is a straight line that has the same slope as the
curve at the given point and goes through the given point. The slope of the tangent line
is equal to the derivative of the function evaluated at the given point.
(a) Start by finding the derivative of y= 2 tan (4x). Recall that yI(tan u)= sec2u ⋅yI.
yI=2 sec2 (4x) ⋅4 → = 8 sec2(4x)
Now evaluate the derivative at x= 𝜋/16.
= 8 sec2(4(𝜋/16))
= 8 sec2 (𝜋/4)
= 8(√2)2
= 16
Use the point-slope form of a line to write an equation for the tangent line and
then solve for y.
y - 2 = 16(x- 𝜋/16)
y = 16x - 𝜋+ 2
Thus, the tangent line for y = 2 tan (4x) at the point (𝜋/16,2) is y=16x - 𝜋+ 2.
(b) To find the tangent line to y = ¼ tan-1(½ x), start by finding its derivative. Recall
that
3.9
1.
2.
f(x)= 9ex
d/dx ex=ex
Find the derivative of y with respect to x of y= ln (x5).
Use the reciprocal rule for logarithms ln xr=r ln x to rewrite the given expression before taking its
derivative.
ln x5 = 5 ln x