Study better. Learn faster. Get the grade you want.
Discover why millions of students use us to learn better.

- StudyBlue
- Indiana
- Indiana University-Bloomington
- Calculus
- Calculus M211
- Kirk
- Midterm Review

Sign up now and start studying these cards for FREE

Derivative

f'(x)=lim(h->0) (f(x+h)-f(x)/h)

Relationship between f(x) and f'(x)

1) f(x) is increasing then f'(x) is decreasing

2) f(x) is decreasing then f'(x) is increasing

3) f(x) has a horizontal tangent then f'(x) crosses x-axis

When is a function not differentiable?

1) There is a discontinuity.

2) There is a kink or corner in the graph.

3) There is a vertical tangent.

Constant Function

d/dx(Constant)=0

Power Rule

d/dx(x^{n})=nx^{n-1}

Constant Multiple Rule

d/dx(c*f(x))=c*d/dx(f(x))

Exponential Functions

f'(x)=a^{x}*lim(h->0) ((a^{h}-1)/h)

Definition of the number e

lim(h->0) (e^{h}-1)/h=1

Derivative of Natural Exponential Function

d/dx(e^{x})=e^{x}

Product Rule

d/dx(f(x)g(x))=f(x)g'(x)+g(x)f'(x)

Quotient Rule

d/dx(f(x)/g(x))=(g(x)f'(x)-f(x)g'(x))/(g(x)^{2})

Chain Rule

F'(x)=f'(g(x))*g'(x)

or

(dy/dx)=(dy/du)(du/dx)

Derivatives of Exponential Functions

d/dx(a^{x})=a^{x}lna

Natural Growth or Decay

y(t)=y_{0}e^{kt}

Total amount of money in account after t years

A(t)=A_{0}(1+r/n)^{nt}

Countinous Compounding

A(t)=A_{0}e^{rt}

Derivatives of Logarithms

f'(log_{a}x)=1/(xlna)

Derivatives of Natural Logs

f'(lnx)=1/x

or

f'(lng(x))=g'(x)/g(x)

Implicit Differentiation

Differentiate both sides and solve for y'

Derivative of a constant base and exponent

f'(a^{b})=0

Derivative of a variable base and constant exponent

d/dx(f(x))^{b}=b(f(x)^{b-1}*f'(x)

Derivative of a constant base and variable exponent

d/dx(a^{g(x)})=a^{g(x)}ln(a)*g'(x)

e as a limit

e=lim(x->0)(1+x)^{(1/x)}

or

e=lim(n->INF)(1+1/n)^{n}

About this deck

Author: Evan B.

Created: 2014-03-06

Updated: 2014-03-06

Size: 23 flashcards

Views: 15

Created: 2014-03-06

Updated: 2014-03-06

Size: 23 flashcards

Views: 15

Simply amazing. The flashcards are smooth, there are
many different types of
studying tools, and there is
a great search engine. I praise
you on the awesomeness.
- Dennis

I have been getting MUCH
better grades on all my tests
for school. Flash cards, notes,
and quizzes are great on here.
Thanks!
- Kathy

I was destroying whole rain forests with my flashcard production, but YOU, StudyBlue, have saved the ozone layer. The earth thanks you.
- Lindsey

This is the greatest app on my phone!! Thanks so much for making it easier to study. This has helped me a lot!
- Tyson

StudyBlue is not sponsored or endorsed by any college, university, or instructor.

© 2014 StudyBlue Inc. All rights reserved.

© 2014 StudyBlue Inc. All rights reserved.