Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus AB : Derivative at a point

Study concepts, example questions & explanations for AP Calculus AB

Create An Account

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #163 : Ap Calculus Ab

Find the equation of the line passing through (-1, 1) and perpendicular to the line tangent to the following function at x=4:

f(x)=x^3

Possible Answers:

y={48}x+49

$y=-\frac{x}{48}+\frac{47}{48}$

$y=\frac{x}{48}+\frac{47}{48}$

y=-{48}x+49

Correct answer:

$y=-\frac{x}{48}+\frac{47}{48}$

Explanation:

To determine the equation of the line, we must first find its slope, which is perpendicular to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

f'(x)=3x^2

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is

f'(4)=48

However, the negative reciprocal of this - the slope of the tangent line - is

$-\frac{1}{48}$

We now can write the equation of the line using point-slope form:

$y-1=-\frac{1}{48}(x+1)$

$y=-\frac{x}{48}+\frac{47}{48}$

Report an Error

Example Question #164 : Ap Calculus Ab

Find the equation of the line passing through (4, 10) and parallel to the tangent line to the following function at x=2:

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

Possible Answers:

$y=-\frac{2}{3}x+16$

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

y=x

$y=\frac{3}{2}x+4$

Correct answer:

$y=\frac{3}{2}x+4$

Explanation:

To determine the equation of the line, we must first find its slope, which is parallel to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

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

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\cdot \frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Now, we find the slope of the tangent line at the given point by plugging this point into the first derivative function:

$f'(2)=\frac{3}{2}$

Now, we use point-slope form to find the equation of the line:

$y-10=\frac{3}{2}(x-4)$

$y=\frac{3}{2}x+4$

Report an Error

Example Question #165 : Ap Calculus Ab

Find the equation of the line passing through (3,1) and parallel to the tangent line to the following function at x=2:

f=x^3-1

Possible Answers:

y=12x-2

y=27

y=12x-35

y=x

None of the other answers

Correct answer:

y=12x-35

Explanation:

f'=3x^2

The following rule was used to find the derivative:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is

3(2^2)=12=m

We now can write the equation of the line using point-slope form:

y-1=12(x-3)

y=12x-35

Report an Error

Example Question #61 : Derivative At A Point

Find the equation of the line that passes through (2,0) and is perpendicular to the tangent line to the following function at x=0:

$f=\ln(2-x)$

Possible Answers:

$y=-\frac{1}{2}x+1$

y=-2x+4

$y=-\frac{1}{2}x+3$

y=2x-4

Correct answer:

y=2x-4

Explanation:

To determine the equation of the line, we must first find its slope, which is perpendicular to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

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

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is

$-\frac{1}{2-0}=-\frac{1}{2}$

However, the line we want has a slope perpendicular to this, so we take the negative reciprocal:

m=2

We now can write the equation of the line using point-slope form:

y-0=2(x-2)

y=2x-4

Report an Error

Example Question #163 : Ap Calculus Ab

Find the slope of the line tangent to the curve of f(x) when x=-2 . Round to the nearest whole number.

f(x)=16e^x+12x^6-48x+13

Possible Answers:

Cannot be determined from the information provided.

2304

Correct answer:

Explanation:

Find the slope of the line tangent to the curve of f(x) when x=-2

f(x)=16e^x+12x^6-48x+13

To find the slope of a tangent line, we need to find our first derivative.

To find our derivative, we need to recall two rules.

$\frac{dy}{dx}e^x=e^x$

And

$\frac{dy}{dx}x^n=nx^{n-1}$

Using these two rules, we can find the derivative of f(x).

Our first term can be derived using our first rule. The derivative of e to the x is just e to the x.

This means that our first term will remain 16e to x.

For our other three terms, we follow the second rule. We will decrease each term's exponent by 1, and then multiply the coefficient by the old exponent.

$f'(x)=16e^x+(6)12x^{6-1}-(1)48x^{1-1}$

Notice that the 13 will drop out. It is a constant term, and as such when we multiply it by it's original exponent (0) it wil be reduced to zero as well.

Clean up the above to get:

$f'(x)=16e^x+72x^{5}-48$

Now, we are almost there. We need to find the slope when x=-2. To do so, plug in -2 for x and solve.

$f'(x)=16e^{-2}+72(-2)^{5}-48\approx -2349.83$

So, our answer is -2350

Report an Error

Example Question #62 : Derivative At A Point

Find the equation of the line passing through the origin and perpendicular to the tangent to the following function at $x=\frac{\pi}{12}$ :

$f=\frac{\cos(3x)}{3}$

Possible Answers:

$y=-\frac{\sqrt{2}x}{2}$

$y=\sqrt{2}x$

y=x

$y=-\sqrt{2}x$

Correct answer:

$y=\sqrt{2}x$

Explanation:

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

The first derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Evaluating the derivative at the given point, we get

$-\sin(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$

Because the slope of the line we want is perpendicular to this line, its slope is the negative reciprocal, or

$\frac{2}{\sqrt{2}}=\sqrt{2}=m$

We can now find the equation of the line using point-slope form:

$y-0=\sqrt{2}(x-0)$

$y=\sqrt{2}x$

Report an Error

Example Question #63 : Derivative At A Point

Find the slope, , of f(x) at x=2 ?

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

Possible Answers:

m=0

m=164

m=136

m=148

m=128

Correct answer:

m=148

Explanation:

The slope of any point on this function can be determined by plugging the point's x-value into the f'(x) , the first derivative of f(x) .

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

f'(x)=10x^4-3x^2

f'(2)=10(2)^4-3(2)^2

=10(16)-3(4)=160-12=148

Report an Error

Example Question #64 : Derivative At A Point

Find m in y = mx + b from the equation, given the point (2,0)

y = (x^2+2x)(-x^3+4)

Possible Answers:

= 48

= -96

= 144

$= -120$

= 100

Correct answer:

$= -120$

Explanation:

To find the tangent line at the given point, we need to first take the derivative of the given function.

To find the derivative we need to use product rule. Product rule states that we take the derivative of the first function and multiply it by the derivative of the second function and then add that with the derivative of the second function multiplied by the given first function. To find the derivative of each separate function we need to use power rule.

f(x)g'(x)+g(x)f'(x)

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent

$n(x^{n-1})$

f(x)= x^2 + 2x ; f'(x)= 2x + 2

g(x)= -x^3 +4 ; g'(x) = -3x^2

Use power rule and we get :

(x^2+2x)(-3x^2)+(-x^3+4)(2x+2)

From here, to find the slope at the given point we plug in "2" for x.

$(2^2 + 2(2))(-3(2^2)) + (- (2^3)+4)(2(2)+2)$

This comes out to equal $-120$

Report an Error

Example Question #22 : Functions, Graphs, And Limits

Which of the following functions contains a removeable discontinuity?

Possible Answers:

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

$f(x)=\frac{1+x^{3}}{1+x}$

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

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

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

Correct answer:

$f(x)=\frac{1+x^{3}}{1+x}$

Explanation:

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at x=a , then the limit as approaches exists, but the value of f(a) does not.

For example, the function $f(x)=\frac{1+x^3}{1+x}$ contains a removeable discontinuity at x=-1 . Notice that we could simplify f(x) as follows:

$f(x)=\frac{1+x^3}{1+x}=\frac{(1+x)(x^2-x+1)}{1+x}=x^{2}-x+1$ , where $x\neq -1$ .

Thus, we could say that $\lim_{x\rightarrow -1}\frac{1+x^3}{1+x}=\lim_{x\rightarrow -1}x^2-x+1=(-1)^2-(-1)+1=3$ .

As we can see, the limit of f(x) exists at x=-1 , even though f(-1) is undefined.

What this means is that f(x) will look just like the parabola with the equation $x^{2}-x+1$ EXCEPT when x=-1 , where there will be a hole in the graph. However, if we were to just define f(-1)=3 , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at x=-1 .

The functions

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

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

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

$f(x)=x^{2}\sin x^{2}$ and $f(x)=\frac{x+1}{1+x^{2}}$ are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

$f(x)=\frac{1+x^{3}}{1+x}$ .

Report an Error

Example Question #1 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&\text{Approximate a forward derivative at }x=4\\&\text{Using the following table:}\\&x&&f(x)\\&1&&5\\&4&&-8\\&7&&15\end{align*}$

Possible Answers:

$\frac{23}{6}$

$-\frac{13}{3}$

$\frac{23}{3}$

$-\frac{8}{3}$

Correct answer:

$\frac{23}{3}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(4)\text{ and }f(7)\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&f'(4)\approx\frac{15-(15)}{3}\\&f'(4)\approx\frac{23}{3}\end{align*}$

Report an Error

View AP Calculus AB Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View AP Calculus AB Tutors

Austin
Certified Tutor

University of Arizona, Bachelor of Science, Physiology.

View AP Calculus AB Tutors

Shakeel
Certified Tutor

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

MCAT Tutors in Washington DC, ACT Tutors in Chicago, LSAT Tutors in Miami, LSAT Tutors in Chicago, ACT Tutors in Los Angeles, Statistics Tutors in Miami, Math Tutors in Boston, Biology Tutors in Chicago, Chemistry Tutors in San Diego, Physics Tutors in San Diego

Popular Courses & Classes

MCAT Courses & Classes in Los Angeles, LSAT Courses & Classes in Washington DC, Spanish Courses & Classes in Houston, SAT Courses & Classes in Atlanta, GMAT Courses & Classes in Seattle, SAT Courses & Classes in New York City, Spanish Courses & Classes in Seattle, SAT Courses & Classes in Phoenix, GMAT Courses & Classes in Philadelphia, MCAT Courses & Classes in Atlanta

Popular Test Prep

ACT Test Prep in Denver, MCAT Test Prep in Washington DC, GRE Test Prep in San Francisco-Bay Area, LSAT Test Prep in San Diego, GMAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Phoenix, ISEE Test Prep in Los Angeles, SSAT Test Prep in Chicago, ACT Test Prep in Boston, SSAT Test Prep in Los Angeles

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Calculus AB : Derivative at a point

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #163 : Ap Calculus Ab

Example Question #164 : Ap Calculus Ab

Example Question #165 : Ap Calculus Ab

Example Question #61 : Derivative At A Point

Example Question #163 : Ap Calculus Ab

Example Question #62 : Derivative At A Point

Example Question #63 : Derivative At A Point

Example Question #64 : Derivative At A Point

Example Question #22 : Functions, Graphs, And Limits

Example Question #1 : Approximate Rate Of Change From Graphs And Tables Of Values

All AP Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: