Derivatives Practice

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1.1State the Meaning of a Derivative

Exam I | Problem 1.1 |

Derivative Meaning

What does $f'(a)$ tell you geometrically about the graph of $y=f(x)$?

Solution

Geometrically, $f'(a)$ is the slope of the tangent line to the graph at $x=a$.

It also describes the function's instantaneous rate of change at that point.

1.2Differentiate by the Definition

Exam I | Problem 1.2 |

Formal Definition

Let

$$ f(x)=x^2. $$

Use the limit definition of the derivative to find $f'(x)$.

Solution

Start with the definition:

f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

Substitute $f(x)=x^2$:

f'(x)=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}

Expand the numerator:

f'(x)=\lim_{h\to 0}\frac{x^2+2xh+h^2-x^2}{h}

Simplify:

f'(x)=\lim_{h\to 0}\frac{2xh+h^2}{h} =\lim_{h\to 0}(2x+h)

Now take the limit:

$$ f'(x)=2x $$

1.3Write the Derivative in Leibniz Notation

Exam I | Problem 1.3 |

Notation

If $y=f(x)$, what is the common Leibniz notation for the derivative with respect to $x$?

Solution

The derivative is written as

\frac{dy}{dx}

This is equivalent to $f'(x)$ when $y=f(x)$.

1.4Differentiate a Power

Exam I | Problem 1.4 |

Power Rule

Find the derivative of

$$ x^7 $$

Solution

Use the power rule:

\frac{d}{dx}[x^n]=nx^{n-1}

\frac{d}{dx}[x^7]=7x^6

1.5Differentiate a Polynomial

Exam I | Problem 1.5 |

Find the derivative of

$$ 3x^4-5x+8 $$

Solution

Differentiate term by term:

\frac{d}{dx}[3x^4]=12x^3

\frac{d}{dx}[-5x]=-5

\frac{d}{dx}[8]=0

So the derivative is

$$ 12x^3-5 $$

1.6Differentiate an Exponential Function

Exam I | Problem 1.6 |

Exponential Functions

Find the derivative of

$$ 2^x $$

Solution

Use the exponential rule:

\frac{d}{dx}[a^x]=a^x\ln a

With $a=2$,

\frac{d}{dx}[2^x]=2^x\ln 2

1.7Differentiate a Trigonometric Sum

Exam I | Problem 1.7 |

Find the derivative of

\sin x-\cos x

Solution

Differentiate each term:

\frac{d}{dx}[\sin x]=\cos x

and

\frac{d}{dx}[-\cos x]=\sin x

So the derivative is

\cos x+\sin x

1.8Differentiate a Natural Logarithm

Exam I | Problem 1.8 |

Logarithmic Functions

Find the derivative of

\ln x

Solution

Use the logarithmic derivative rule:

\frac{d}{dx}[\ln x]=\frac{1}{x} \quad \text{for } x>0

1.9Use the Product Rule

Exam I | Problem 1.9 |

Find the derivative of

$$ x^2e^x $$

Solution

Use the product rule:

\frac{d}{dx}[fg]=f'g+fg'

Let $f=x^2$ and $g=e^x$. Then $f'=2x$ and $g'=e^x$.

\frac{d}{dx}[x^2e^x]=2xe^x+x^2e^x

1.10Use the Quotient Rule

Exam I | Problem 1.10 |

Quotient Rule

Find the derivative of

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

and simplify your answer.

Solution

Use the quotient rule:

\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f'g-fg'}{g^2}

Let $f=x^2+1$ and $g=x$. Then $f'=2x$ and $g'=1$.

\frac{d}{dx}\left[\frac{x^2+1}{x}\right] = \frac{(2x)(x)-(x^2+1)(1)}{x^2}

Simplify:

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

Difficulty

Integrated

2.1Apply the Chain Rule to a Power

Exam II | Problem 2.1 |

Find the derivative of

$$ (3x^2+1)^5 $$

Solution

Use the chain rule. Differentiate the outside first and then multiply by the derivative of the inside:

\frac{d}{dx}[(3x^2+1)^5] =5(3x^2+1)^4(6x)

So the derivative is

$$ 30x(3x^2+1)^4 $$

2.2Apply the Chain Rule to a Trigonometric Function

Exam II | Problem 2.2 |

Find the derivative of

\sin(x^3)

Solution

Treat the inside as the inner function. The derivative of $\sin u$ is $\cos u$, and the derivative of $x^3$ is $3x^2$.

\frac{d}{dx}[\sin(x^3)] =\cos(x^3)\cdot 3x^2

Thus,

3x^2\cos(x^3)

2.3Differentiate a Logarithm of a Composite Function

Exam II | Problem 2.3 |

Find the derivative of

\ln(2x^2-7x+4)

Solution

Use the chain rule with $u=2x^2-7x+4$.

The derivative of $\ln u$ is $\frac{1}{u}u'$.

Here,

$$ u'=4x-7 $$

\frac{d}{dx}[\ln(2x^2-7x+4)] =\frac{4x-7}{2x^2-7x+4}

2.4Differentiate an Inverse Trig Function

Exam II | Problem 2.4 |

Find the derivative of

\arctan(3x)

Solution

Use the inverse trig rule and the chain rule:

\frac{d}{dx}[\arctan u]=\frac{u'}{1+u^2}

Here $u=3x$, so $u'=3$.

Therefore,

\frac{d}{dx}[\arctan(3x)] =\frac{3}{1+(3x)^2} =\frac{3}{1+9x^2}

2.5Implicit Differentiation with Mixed Terms

Exam II | Problem 2.5 |

Differentiate implicitly and solve for $\frac{dy}{dx}$:

$$ x^2+xy+y^2=7 $$

Solution

Differentiate each term with respect to $x$:

2x+\frac{d}{dx}[xy]+\frac{d}{dx}[y^2]=0

Use the product rule on $xy$ and the chain rule on $y^2$:

2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0

Group the $\frac{dy}{dx}$ terms:

\left(x+2y\right)\frac{dy}{dx}=-(2x+y)

\frac{dy}{dx}=-\frac{2x+y}{x+2y}

2.6Find Velocity and Acceleration

Exam II | Problem 2.6 |

Let

$$ s(t)=t^3-6t^2+9t $$

Find the velocity $v(t)$, the acceleration $a(t)$, and the value of $a(2)$.

Solution

Velocity is the first derivative of position:

$$ v(t)=s'(t)=3t^2-12t+9 $$

Acceleration is the derivative of velocity:

$$ a(t)=v'(t)=6t-12 $$

Now evaluate at $t=2$:

$$ a(2)=6(2)-12=0 $$

2.7Estimate a Square Root with Linearization

Exam II | Problem 2.7 |

Linear Approximation

Use the linear approximation of $f(x)=\sqrt{x}$ at $x=16$ to estimate $\sqrt{16.2}$.

Solution

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

$$ f(16)=4 $$

and

f'(x)=\frac{1}{2\sqrt{x}}

f'(16)=\frac{1}{8}

The linearization at $x=16$ is

L(x)=4+\frac{1}{8}(x-16)

Evaluate at $x=16.2$:

L(16.2)=4+\frac{1}{8}(0.2)=4.025

\sqrt{16.2}\approx 4.025

2.8Find Where a Function Increases

Exam II | Problem 2.8 |

Suppose

$$ f'(x)=3(x-2)(x+1). $$

On which intervals is $f$ increasing and on which intervals is it decreasing?

Solution

First find the critical numbers from $f'(x)=0$:

$$ 3(x-2)(x+1)=0 $$

So the critical numbers are $x=-1$ and $x=2$.

Check the sign of $f'(x)$ on each interval:

If $x<-1$, then both factors are negative, so $f'(x)>0$.
If $-1<x<2$, then one factor is negative and one is positive, so $f'(x)<0$.
If $x>2$, then both factors are positive, so $f'(x)>0$.

Therefore, $f$ is increasing on

(-\infty,-1)\cup(2,\infty)

and decreasing on

$$ (-1,2). $$

Difficulty

Applied

3.1Related Rates for the Area of a Circle

Final | Problem 3.1 |

Related Rates

A circle's radius is increasing at a rate of $0.5$ cm/s. How fast is the area changing when the radius is $10$ cm?

Solution

Use the area formula

A=\pi r^2

Differentiate with respect to time:

\frac{dA}{dt}=2\pi r\frac{dr}{dt}

Substitute $r=10$ and $\frac{dr}{dt}=0.5$:

\frac{dA}{dt}=2\pi(10)(0.5)=10\pi

So the area is changing at

10\pi \text{ cm}^2/\text{s}

3.2Tangent Line from an Implicit Curve

Final | Problem 3.2 |

Find the equation of the tangent line to

$$ x^2+y^2=34 $$

at the point $(3,5)$.

Solution

Differentiate implicitly:

2x+2y\frac{dy}{dx}=0

Solve for the derivative:

\frac{dy}{dx}=-\frac{x}{y}

At $(3,5)$, the slope is

\frac{dy}{dx}=-\frac{3}{5}

Use point-slope form:

y-5=-\frac{3}{5}(x-3)

3.3Maximize the Area of a Rectangle

Final | Problem 3.3 |

Optimization

A rectangle has perimeter $40$ m. What dimensions give the maximum area?

Solution

Let the side lengths be $x$ and $y$.

The perimeter condition is

$$ 2x+2y=40 $$

$$ y=20-x $$

The area is

$$ A(x)=x(20-x)=20x-x^2 $$

Differentiate:

$$ A'(x)=20-2x $$

Set the derivative equal to zero:

20-2x=0 \quad\Rightarrow\quad x=10

Then

$$ y=20-10=10 $$

So the rectangle with maximum area is a $10$ m by $10$ m square.

3.4Estimate Measurement Error with Differentials

Final | Problem 3.4 |

A circular disk has radius $10$ cm, and the radius measurement may be off by about $0.05$ cm. Use differentials to estimate the possible error in the area.

Solution

For a circle,

A=\pi r^2

dA=2\pi r\,dr

Substitute $r=10$ and $dr=0.05$:

dA=2\pi(10)(0.05)=\pi

So the area may be off by about

\pi \text{ cm}^2

which is approximately $3.14\text{ cm}^2$.

3.5Take One Newton Step

Final | Problem 3.5 |

Newton's Method

Use one step of Newton's method to approximate the root of

$$ x^3-2=0 $$

starting from $x_0=1$.

Solution

Let

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

Then

$$ f'(x)=3x^2 $$

Newton's method gives

x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

With $x_0=1$:

x_1=1-\frac{1^3-2}{3(1)^2}

x_1=1-\frac{-1}{3}=1+\frac{1}{3}=\frac{4}{3}

So the next approximation is

\frac{4}{3}

Difficulty

Challenge

4.1Show an Absolute Value Function Is Not Differentiable

Final | Problem 4.1 |

Show that

$$ f(x)=|x| $$

is continuous at $x=0$ but not differentiable there.

Solution

First check continuity:

\lim_{x\to 0}|x|=0=f(0)

so $f$ is continuous at $0$.

Now check differentiability using one-sided slopes. For $x>0$, $f(x)=x$, so the right-hand derivative at $0$ is

\lim_{x\to 0^+}\frac{|x|-|0|}{x-0} =\lim_{x\to 0^+}\frac{x}{x}=1

For $x<0$, $f(x)=-x$, so the left-hand derivative at $0$ is

\lim_{x\to 0^-}\frac{|x|-|0|}{x-0} =\lim_{x\to 0^-}\frac{-x}{x}=-1

The one-sided derivatives are not equal, so $f'(0)$ does not exist.

4.2Classify a Cubic Using Derivatives

Final | Problem 4.2 |

For

$$ f(x)=x^3-3x, $$

find the critical points, classify them, and identify an inflection point.

Solution

Differentiate once:

$$ f'(x)=3x^2-3=3(x-1)(x+1) $$

So the critical points occur at

x=-1 \quad \text{and} \quad x=1

Differentiate again:

$$ f''(x)=6x $$

At $x=-1$,

$$ f''(-1)=-6<0 $$

so $x=-1$ is a local maximum.

At $x=1$,

$$ f''(1)=6>0 $$

so $x=1$ is a local minimum.

For an inflection point, check where concavity changes. Since

$$ f''(x)=6x $$

changes sign at $x=0$, there is an inflection point at $x=0$.

4.3Maximize Area with a Fence and a Wall

Final | Problem 4.3 |

Optimization

A rectangular pen is built against a straight wall, so only three sides need fencing. If $24$ m of fencing are available, what dimensions maximize the area?

Solution

Let $x$ be the two equal sides perpendicular to the wall, and let $y$ be the side parallel to the wall.

The fencing constraint is

$$ 2x+y=24 $$

$$ y=24-2x $$

The area is

$$ A(x)=xy=x(24-2x)=24x-2x^2 $$

Differentiate:

$$ A'(x)=24-4x $$

Set the derivative equal to zero:

24-4x=0 \quad\Rightarrow\quad x=6

Then

$$ y=24-2(6)=12 $$

So the maximum area occurs for dimensions $6$ m by $6$ m by $12$ m, with the $12$ m side against the wall.

4.4Find a Tangent Line on a Mixed Implicit Curve

Final | Problem 4.4 |

For the curve

$$ x^2+xy+y^2=7, $$

find $\frac{dy}{dx}$ and the equation of the tangent line at $(2,1)$.

Solution

Differentiate implicitly:

2x+\frac{d}{dx}[xy]+2y\frac{dy}{dx}=0

Use the product rule on $xy$:

2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0

Group the derivative terms:

(x+2y)\frac{dy}{dx}=-(2x+y)

\frac{dy}{dx}=-\frac{2x+y}{x+2y}

At $(2,1)$,

\frac{dy}{dx}=-\frac{2(2)+1}{2+2(1)}=-\frac{5}{4}

Use point-slope form:

y-1=-\frac{5}{4}(x-2)