Calculus 1:Differentiation

/30

Test your understanding of fundamental differentiation concepts in Calculus 1. This quiz covers rules of differentiation, derivatives of basic functions, applications like slope and rate of change, and problem-solving using first principles.

1 / 30

The derivative of

$f (x) = 3 x^{4} - 2 x^{2} + 5$

is:

12 x^{3} - 4 x

12 x^{3} - 2 x

3 x^{3} - 2 x

6 x^{3} - 4 x

Apply the power rule:

$\frac{d}{d x} (x^{n}) = n x^{n - 1}$

2 / 30

$f (x) = \sqrt{x}$

, then

$f^{'} (x)$

is:

\frac{1}{\sqrt{x}}

2 \sqrt{x}

\frac{1}{2 \sqrt{x}}

\sqrt{x}

Rewrite

$\sqrt{x} = x^{1 / 2}$

, then apply the power rule.

3 / 30

The derivative of

$f (x) = \frac{1}{x^{2}}$

is:

\frac{2}{x^{3}}

- \frac{2}{x^{3}}

- \frac{1}{x^{3}}

\frac{1}{x^{3}}

Rewrite

$\frac{1}{x^{2}} = x^{- 2}$

, then differentiate.

4 / 30

$f (x) = e^{3 x}$

, then

$f^{'} (x)$

is:

e^{3 x}

3 e^{x}

\frac{e^{3 x}}{3}

3 e^{3 x}

Use the chain rule:

$\frac{d}{d x} e^{u} = e^{u} \cdot u^{'}$

5 / 30

The derivative of

$f (x) = \ln (2 x)$

is:

\frac{1}{x}

\frac{2}{x}

\frac{1}{2 x}

\ln (2)

$\frac{d}{d x} \ln (u) = \frac{u^{'}}{u}$

. Here,

$u = 2 x$

6 / 30

$f (x) = \sin (4 x)$

, then

$f^{'} (x)$

is:

\cos (4 x)

4 \cos (4 x)

- 4 \cos (4 x)

- \cos (4 x)

Chain rule:

$\frac{d}{d x} \sin (u) = \cos (u) \cdot u^{'}$

7 / 30

The derivative of

$f (x) = 5$

is:

0

5

5 x

Undefined

8 / 30

$f (x) = x^{2} \sin (x)$

, then

$f^{'} (x)$

is:

2 x \cos (x)

x^{2} \cos (x)

2 x \sin (x) + x^{2} \cos (x)

2 x \sin (x)

Product rule:

$\frac{d}{d x} (u v) = u^{'} v + u v^{'}$

9 / 30

The derivative of

$f (x) = \tan (x)$

is:

\cot (x)

- \csc^{2} (x)

\tan^{2} (x)

\sec^{2} (x)

Standard derivative of

$\tan (x)$

10 / 30

$f (x) = \frac{x + 1}{x - 1}$

, then

$f^{'} (x)$

is:

\frac{- 2}{(x - 1)^{2}}

\frac{2}{(x - 1)^{2}}

\frac{1}{(x - 1)^{2}}

\frac{x - 1}{(x + 1)^{2}}

Quotient rule:

$\frac{d}{d x} (\frac{u}{v}) = \frac{u^{'} v - u v^{'}}{v^{2}}$

11 / 30

The critical points of

$f (x) = x^{3} - 3 x^{2}$

occur at:

x = 1

and

x = 3

x = - 1

and

x = 1

x = 0

and

x = 2

x = 0

and

x = 1

Find

$f^{'} (x) = 3 x^{2} - 6 x$

, set

$f^{'} (x) = 0$

12 / 30

The function

$f (x) = x^{3} - 6 x^{2} + 9 x$

has a local maximum at:

x = 3

x = 1

x = 0

x = 2

Find

$f^{'} (x) = 3 x^{2} - 12 x + 9$

, solve

$f^{'} (x) = 0$

, and use the second derivative test.

13 / 30

The tangent line to

$y = x^{2}$

$x = 1$

is:

y = x + 1

y = 2 x + 1

y = x - 1

y = 2 x - 1

Slope

$m = f^{'} (1) = 2$

, point

$(1, 1)$

14 / 30

The acceleration of a particle with position

$s (t) = t^{3} - 6 t^{2}$

is:

3 t^{2} - 12 t

6 t - 12

t^{3} - 6 t

6 t

15 / 30

The function

$f (x) = x^{4} - 4 x^{3}$

is concave up where:

x < 2

x > 0

x > 2

x < 0

Find

$f^{''} (x) = 12 x^{2} - 24 x$

, solve

$f^{''} (x) > 0$

16 / 30

The linear approximation of

$f (x) = \sqrt{x}$

near

$x = 4$

is:

L (x) = 2 + \frac{1}{4} (x - 4)

L (x) = 2 + \frac{1}{2} (x - 4)

L (x) = 4 + \frac{1}{2} (x - 2)

L (x) = 2 + 4 (x - 4)

$L (x) = f (a) + f^{'} (a) (x - a)$

, where

$a = 4$

17 / 30

The absolute maximum of

$f (x) = x^{3} - 12 x$

$[0, 3]$

is:

- 16

0

3

9

Evaluate

$f (x)$

at critical points

$x = 2$

and endpoints

$x = 0, 3$

18 / 30

The derivative

$\frac{d y}{d x}$

for

$y^{2} = x^{3}$

is:

\frac{2 y}{3 x^{2}}

3 x^{2} y

\frac{3 x^{2}}{2 y}

\frac{3 x}{2 y}

Differentiate implicitly:

$2 y \frac{d y}{d x} = 3 x^{2}$

19 / 30

The elasticity of demand

$E$

for

$p = 10 - 2 q$

is:

E = \frac{10}{p}

E = \frac{p}{10 - p}

E = \frac{2 p}{q}

E = \frac{q}{p}

Elasticity

$E = ∣ \frac{p}{q} \cdot \frac{d q}{d p} ∣$

20 / 30

The derivative of

$f (x) = \arcsin (x)$

is:

\frac{1}{\sqrt{1 - x^{2}}}

\frac{- 1}{\sqrt{1 - x^{2}}}

\frac{1}{1 + x^{2}}

\frac{- 1}{1 + x^{2}}

21 / 30

The second derivative of

$f (x) = x^{3} - 2 x$

is:

3 x^{2} - 2

6 x^{2}

6

6 x

First derivative

$f^{'} (x) = 3 x^{2} - 2$

, second derivative

$f^{''} (x) = 6 x$

22 / 30

$f^{''} (x) = 12 x^{2}$

, then

$f (x)$

could be:

x^{4} + C x + D

4 x^{3} + C

12 x + C

6 x^{2} + C

Integrate twice:

$f^{'} (x) = 4 x^{3} + C$

$f (x) = x^{4} + C x + D$

23 / 30

The third derivative of

$f (x) = \sin (x)$

is:

\cos (x)

\sin (x)

- \cos (x)

- \sin (x)

$f^{'} (x) = \cos (x)$

$f^{''} (x) = - \sin (x)$

$f^{'''} (x) = - \cos (x)$

24 / 30

The second derivative of

$y = e^{2 x}$

is:

2 e^{2 x}

4 e^{2 x}

e^{2 x}

4 e^{x}

$y^{'} = 2 e^{2 x}$

$y^{''} = 4 e^{2 x}$

25 / 30

If $f^{''} (x) = 0$ for all $x$ , then $f (x)$ is:

A linear function

A quadratic function

A constant function

A cubic function

The general solution is $f (x) = C x + D$ .

26 / 30

The second derivative of

$f (x) = \ln (x)$

is:

\frac{1}{x^{2}}

- \frac{1}{x^{2}}

\frac{1}{x}

- \ln (x)

$f^{'} (x) = \frac{1}{x}$

$f^{''} (x) = - \frac{1}{x^{2}}$

27 / 30

The third derivative of

$f (x) = x^{5}$

is:

60 x^{2}

20 x^{3}

120 x

5 x^{4}

$f^{'} (x) = 5 x^{4}$

$f^{''} (x) = 20 x^{3}$

$f^{'''} (x) = 60 x^{2}$

28 / 30

$f^{''} (x) = 6 x$

, then

$f (x)$

could be:

3 x^{2} + C

6 x + C

x^{3} + C x + D

x^{2} + C

Integrate twice:

$f^{'} (x) = 3 x^{2} + C$

$f (x) = x^{3} + C x + D$

29 / 30

The second derivative of

$f (x) = \cos (3 x)$

is:

9 \cos (3 x)

- 3 \sin (3 x)

3 \sin (3 x)

- 9 \cos (3 x)

$f^{'} (x) = - 3 \sin (3 x)$

$f^{''} (x) = - 9 \cos (3 x)$

30 / 30

The second derivative of

$y = \frac{1}{x}$

is:

\frac{2}{x^{3}}

- \frac{1}{x^{2}}

\frac{1}{x^{3}}

- \frac{2}{x^{3}}

$y^{'} = - \frac{1}{x^{2}}$

$y^{''} = \frac{2}{x^{3}}$

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