Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Definition of Derivative

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #51 : Definition Of Derivative

Find the derivative of f(x)=sinx using the definition of the derivative.

Possible Answers:

f'(x)=cos(x)

f'(x)=-sin(x)

f'(x)=sin(x)

f'(x)=-cos(x)

Correct answer:

f'(x)=cos(x)

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=sinx

f(x+h)=sin(x+h)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{sin(x+h)-sin(x)}{h}$

Substituting in the trigonometric identity

sin(A+B)=sinAcosB+cosAsinB

$f'(x)=\lim_{h\rightarrow 0}\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{sin(x)(cos(h)-1)+cos(x)sin(h)}{h}$

Then using the identities

$\lim_{h\rightarrow 0}\frac{cos(h)-1}{h}=0$ and $\lim_{h\rightarrow 0}\frac{sin(h)}{h}=1$

$f'(x)=\lim_{h\rightarrow 0}\left [sin(x)\frac{(cos(h)-1)}{h}+cos(x)\frac{sin(h)}{h} \right ]$

$f'(x)=sin(x)\lim_{h\rightarrow 0}\left [\frac{(cos(h)-1)}{h} \right ]+cos(x)\lim_{h\rightarrow 0}\left [\frac{sin(h)}{h} \right ]$

f'(x)=sin(x)*0+cos(x)*1

f'(x)=cos(x)

Report an Error

Example Question #52 : Derivative Review

Find the derivative of f(x)=cosx using the definition of the derivative.

Possible Answers:

f'(x)=-cos(x)

f'(x)=-sin(x)

f'(x)=sin(x)

f'(x)=cos(x)

Correct answer:

f'(x)=sin(x)

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=cosx

f(x+h)=cos(x+h)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{cos(x+h)-cos(x)}{h}$

Substituting in the trigonometric identity

cos(A-B)=cosAcosB+sinAsinB

$f'(x)=\lim_{h\rightarrow 0}\frac{cos(x)cos(h)+sin(x)sin(h)-cos(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{cos(x)(cos(h)-1)+sin(x)sin(h)}{h}$

Then using the identities

$\lim_{h\rightarrow 0}\frac{cos(h)-1}{h}=0$ and $\lim_{h\rightarrow 0}\frac{sin(h)}{h}=1$

$f'(x)=\lim_{h\rightarrow 0}\left [cos(x)\frac{(cos(h)-1)}{h}+sin(x)\frac{sin(h)}{h} \right ]$

$f'(x)=cos(x)\lim_{h\rightarrow 0}\left [\frac{(cos(h)-1)}{h} \right ]+sin(x)\lim_{h\rightarrow 0}\left [\frac{sin(h)}{h} \right ]$

f'(x)=cos(x)*0+sin(x)*1

f'(x)=sin(x)

Report an Error

Example Question #53 : Derivative Review

Find the derivative of $f(x)=\frac{x+3}{x-2}$ using the definition of the derivative.

Possible Answers:

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

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

$f'(x)=\frac{-5}{x^{2}-4x+4}$

f'(x)=1

Correct answer:

$f'(x)=\frac{-5}{x^{2}-4x+4}$

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=\frac{x+3}{x-2}$

$f(x+h)=\frac{x+h+3}{x+h-2}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{x+h+3}{x+h-2}-\frac{x+3}{x-2}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{x+h+3}{x+h-2}-\frac{x+3}{x-2} \right )$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{(x+h+3)(x-2)-(x+3)(x+h-2)}{(x+h-2)(x-2)}\right )$

$f'(x)=\lim_{h\rightarrow 0} \frac{(x+h+3)(x-2)-(x+3)(x+h-2)}{h(x+h-2)(x-2)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(x^{2}+hx+3x-2x-2h-6)-(x^{2}+hx-2x+3x+3h-6)}{h(x^{2}+hx-2x-2x-2h+4)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{x^{2}+hx+x-2h-6-x^{2}-hx-x-3h+6}{h(x^{2}+hx-4x-2h+4)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{-5h}{h(x^{2}+hx-4x-2h+4)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{-5}{(x^{2}+hx-4x-2h+4)}$

$f'(x)=\frac{-5}{x^{2}+(0)x-4x-2(0)+4}$

$f'(x)=\frac{-5}{x^{2}-4x+4}$

Report an Error

Example Question #54 : Derivative Review

Find the derivative of $f(x)=\frac{5x^2-1}{x+3}$ using the definition of the derivative.

Possible Answers:

f'(x)={10x}

$f'(x)=\frac{5x^2+30x+1}{x^2+6x+9}$

$f'(x)=\frac{x^2+6x-2}{x^2+3x-7}$

$f'(x)=\frac{5x^2-1}{x+3}$

Correct answer:

$f'(x)=\frac{5x^2+30x+1}{x^2+6x+9}$

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=\frac{5x^2-1}{x+3}$

$f(x+h)=\frac{5(x+h)^2-1}{x+h+3}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{5(x+h)^2-1}{x+h+3}-\frac{5(x)^2-1}{x+3}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{5(x+h)^2-1}{x+h+3}-\frac{5(x)^2-1}{x+3} \right )$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5(x+h)^{2}-1)(x+3)-(5x^{2}-1)(x+h+3)}{h(x+h+3)(x+3)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5(x^2+2xh+h^2)-1)(x+3)-(5x^{2}-1)(x+h+3)}{h(x+h+3)(x+3)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5x^2+10xh+5h^2-1)(x+3)-(5x^{2}-1)(x+h+3)}{h(x+h+3)(x+3)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5x^3+10x^2h+5h^2x-x+15x^2+30xh+15h^2-3)-(5x^3+5hx^2+15x^2-x-h-3)}{h(x^2+hx+3x+3x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{5x^3+10x^2h+5h^2x-x+15x^2+30xh+15h^2-3-5x^3-5hx^2-15x^2+x+h+3}{h(x^2+hx+3x+3x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{5x^2h+5h^2x+30xh+15h^2+h}{h(x^2+hx+6x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{h(5x^2+5hx+30x+15h+1)}{h(x^2+hx+6x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{5x^2+5hx+30x+15h+1}{x^2+hx+6x+3h+9}$

$f'(x)=\frac{5x^2+5(0)x+30x+15(0)+1}{x^2+(0)x+6x+3(0)+9}$

$f'(x)=\frac{5x^2+30x+1}{x^2+6x+9}$

Report an Error

Example Question #55 : Derivative Review

Find the derivative of $f(x)=x^{3}-3x^{2}-6x+10$ using the definition of the derivative.

Possible Answers:

$f'(x)=x^{3}-3x^{2}-6x+10$

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

f'(x)=3x^2-6x-6

f(x)=10

Correct answer:

f'(x)=3x^2-6x-6

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=x^{3}-3x^{2}-6x+10$

$f(x+h)=(x+h)^{3}-3(x+h)^{2}-6(x+h)+10$

f(x+h)=(x^3+3xh^2+3x^2h+h^3)-3(x^2+2xh+h^2)-6(x+h)+10

f(x+h)=x^3+3xh^2+3x^2h+h^3-3x^2-6xh-3h^2-6x-6h+10

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+3xh^2+3x^2h+h^3-3x^2-6xh-3h^2-6x-6h+10-(x^{3}-3x^{2}-6x+10)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+3xh^2+3x^2h+h^3-3x^2-6xh-3h^2-6x-6h+10-x^{3}+3x^{2}+6x-10}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{3xh^2+3x^2h+h^3-6xh-3h^2-6h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h(3xh+3x^2+h^2-6x-3h-6)}{h}$

$f'(x)=\lim_{h\rightarrow 0}3xh+3x^2+h^2-6x-3h-6$

f'(x)=3x(0)+3x^2+(0)^2-6x-3(0)-6

f'(x)=3x^2-6x-6

Report an Error

Example Question #51 : Derivative Review

Find the derivative of $f(x)=\sqrt{x^{2}-4x}$ using the definition of the derivative.

Possible Answers:

$f'(x)=\frac{x-2}{\sqrt{x^2-4x}}$

$f'(x)=\sqrt{x^{2}-4x}$

$f'(x)=\frac{2x-4}{\sqrt{x^2-4x}}$

$f'(x)=\sqrt{2x-4}$

Correct answer:

$f'(x)=\frac{x-2}{\sqrt{x^2-4x}}$

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=\sqrt{x^{2}-4x}$

$f(x+h)=\sqrt{(x+h)^{2}-4(x+h)}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{(x+h)^{2}-4(x+h)}-\sqrt{x^{2}-4x}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{x^2+2xh+h^2-4x-4h}-\sqrt{x^{2}-4x}}{h}$

Next we will multiply the numerator and denominator by the complex conjugate of the numerator

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{x^2+2xh+h^2-4x-4h}-\sqrt{x^{2}-4x}}{h}*\frac{\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x}}{\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x}}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\left (\sqrt{x^2+2xh+h^2-4x-4h} \right )^2-\left (\sqrt{x^{2}-4x} \right )^2}{h*(\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{ x^2+2xh+h^2-4x-4h-(x^{2}-4x)}{h*(\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{ 2xh+h^2-4h}{h*(\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h*( 2x+h-4)}{h*(\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{( 2x+h-4)}{\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x}}$

$f'(x)=\frac{2x+(0)-4}{\sqrt{x^2+2x(0)+(0)^2-4x-4(0)}+\sqrt{x^{2}-4x}}$

$f'(x)=\frac{2x-4}{\sqrt{x^2-4x}+\sqrt{x^{2}-4x}}$

$f'(x)=\frac{2(x-2)}{2\sqrt{x^2-4x}}$

$f'(x)=\frac{x-2}{\sqrt{x^2-4x}}$

Report an Error

Example Question #51 : Definition Of Derivative

Find the derivative of f(x)=(x-7)(x-9) using the definition of the derivative.

Possible Answers:

f'(x)=x^2

f'(x)=2x-16

f'(x)=5x-2

f'(x)=7

Correct answer:

f'(x)=2x-16

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=(x-7)(x-9)

f(x)=x^2-7x-9x+63

f(x)=x^2-16x+63

f(x+h)=(x+h-7)(x+h-9)

f(x+h)=(x^2+hx-7x+hx+h^2-7h-9x-9h+63)

f(x+h)=(x^2+2hx-16x+h^2-16h+63)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{(x^2+2hx-16x+h^2-16h+63)-(x^2-16x+63)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^2+2hx-16x+h^2-16h+63-x^2+16x-63}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2hx+h^2-16h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h*(2x+h-16)}{h}$

$f'(x)=\lim_{h\rightarrow 0}(2x+h-16)$

f'(x)=2x+0-16

f'(x)=2x-16

Report an Error

Example Question #58 : Derivative Review

Find the derivative of $f(x)=\sqrt{x-7}$ using the definition of the derivative.

Possible Answers:

$f'(x)=\sqrt{x-7}$

$f'(x)=\frac{1}{2\sqrt{x-7}}$

f(x)=1

$f'(x)=\frac{x-7}{\sqrt{x-7}}$

Correct answer:

$f'(x)=\frac{1}{2\sqrt{x-7}}$

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=\sqrt{x-7}$

$f(x+h)=\sqrt{x+h-7}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{x+h-7}-\sqrt{x-7}}{h}$

Next we will multiply the numerator and denominator by the complex conjugate of the numerator

$f'(x)=\lim_{h\rightarrow 0}\frac{\sqrt{x+h-7}-\sqrt{x-7}}{h}*\frac{\sqrt{x+h-7}+\sqrt{x-7}}{\sqrt{x+h-7}+\sqrt{x-7}}$

$f'(x)=\lim_{h\rightarrow 0}\frac{(\sqrt{x+h-7})^2-(\sqrt{x-7})^2}{h(\sqrt{x+h-7}+\sqrt{x-7})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x+h-7-(x-7)}{h(\sqrt{x+h-7}+\sqrt{x-7})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x+h-7-x+7}{h(\sqrt{x+h-7}+\sqrt{x-7})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h}{h(\sqrt{x+h-7}+\sqrt{x-7})}$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h-7}+\sqrt{x-7}}$

$f'(x)=\frac{1}{\sqrt{x+(0)-7}+\sqrt{x-7}}$

$f'(x)=\frac{1}{2\sqrt{x-7}}$

Report an Error

Example Question #59 : Derivative Review

Find the derivative of f(x)=(2x+3)(x-16) using the definition of the derivative.

Possible Answers:

f'(x)=1

f'(x)=2x^2

f'(x)=4x-29

f'(x)=3x-16

Correct answer:

f'(x)=4x-29

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=(2x+3)(x-16)

f(x)=2x^2+3x-32x-48

f(x)=2x^2-29x-48

f(x+h)=(2(x+h)+3)(x+h-16)

f(x+h)=(2x+2h+3)(x+h-16)

f(x+h)=(2x^2+2hx+3x+2xh+2h^2+3h-32x-32h-48)

f(x+h)=2x^2+2h^2+4hx-29x-29h-48

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{(2x^2+2h^2+4hx-29x-29h-48)-(2x^2-29x-48)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2x^2+2h^2+4hx-29x-29h-48-2x^2+29x+48}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2h^2+4hx-29h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h(2h+4x-29)}{h}$

$f'(x)=\lim_{h\rightarrow 0}(2h+4x-29)$

f'(x)=(2(0)+4x-29)

f'(x)=4x-29

Report an Error

Example Question #60 : Derivative Review

Find the derivative of f(x)=(x+2)(x+3)(x+4) using the definition of the derivative.

Possible Answers:

f'(x)=-x^2+7x +37

f'(x)=4x^2+7x +49

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

f'(x)=3x^2+18x +26

Correct answer:

f'(x)=3x^2+18x +26

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

f(x)=(x+2)(x+3)(x+4)

f(x)=(x^2+5x+6)(x+4)

f(x)=x^3+5x^2+6x +4x^2+20x+24

f(x)=x^3+9x^2+26x+24

f(x+h)=(x+h+2)(x+h+3)(x+h+4)

f(x+h)=(x^2+hx+2x +hx+h^2+2h +3x+3h+6)(x+h+4)

f(x+h)=(x^2+2hx+5x+h^2+5h+6)(x+h+4)

f(x+h)=(x^3+2hx^2+5x^2+h^2x+5hx+6x +x^2h+2h^2x+5xh+h^3+5h^2+6h +4x^2+8hx+20x+4h^2+20h+24)

f(x+h)=(x^3+h^3+3x^2h+3h^2x+9x^2+9h^2+18hx +26x+26h +24)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{(x^3+h^3+3x^2h+3h^2x+9x^2+9h^2+18hx +26x+26h +24)-(x^3+9x^2+26x+24)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+h^3+3x^2h+3h^2x+9x^2+9h^2+18hx +26x+26h +24-x^3-9x^2-26x-24}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h^3+3x^2h+3h^2x+9h^2+18hx +26h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h*(h^2+3x^2+3hx+9h+18x +26)}{h}$

$f'(x)=\lim_{h\rightarrow 0}(h^2+3x^2+3hx+9h+18x +26)$

f'(x)=(0)^2+3x^2+3(0)x+9(0)+18x +26

f'(x)=3x^2+18x +26

Report an Error

View Calculus 2 Tutors

Gregory
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Chemical Engineering. Carnegie Mellon University, Doctor of P...

View Calculus 2 Tutors

Muhammad
Certified Tutor

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

View Calculus 2 Tutors

Dossevi
Certified Tutor

cole Ste Genevive Versailles, Bachelor of Science, Mathematics. Institut National Polytechnique de Grenoble, Master of Scienc...

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

French Tutors in San Diego, GMAT Tutors in San Diego, ISEE Tutors in Houston, English Tutors in San Francisco-Bay Area, Spanish Tutors in Miami, English Tutors in Atlanta, LSAT Tutors in Atlanta, Algebra Tutors in Philadelphia, GRE Tutors in Seattle, LSAT Tutors in Seattle

Popular Courses & Classes

Spanish Courses & Classes in Atlanta, SAT Courses & Classes in Philadelphia, MCAT Courses & Classes in Washington DC, ACT Courses & Classes in Seattle, SSAT Courses & Classes in Atlanta, GMAT Courses & Classes in Boston, ISEE Courses & Classes in Miami, LSAT Courses & Classes in Philadelphia, ACT Courses & Classes in Atlanta, ACT Courses & Classes in New York City

Popular Test Prep

GMAT Test Prep in Boston, SSAT Test Prep in Houston, LSAT Test Prep in Phoenix, GRE Test Prep in Washington DC, ISEE Test Prep in Washington DC, LSAT Test Prep in Houston, ACT Test Prep in New York City, GMAT Test Prep in Los Angeles, ISEE Test Prep in Phoenix, MCAT Test Prep in Seattle

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

Calculus 2 : Definition of Derivative

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #51 : Definition Of Derivative

Example Question #52 : Derivative Review

Example Question #53 : Derivative Review

Example Question #54 : Derivative Review

Example Question #55 : Derivative Review

Example Question #51 : Derivative Review

Example Question #51 : Definition Of Derivative

Example Question #58 : Derivative Review

Example Question #59 : Derivative Review

Example Question #60 : Derivative Review

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: