Chain Rule
Limits
Integration
Derivatives
Theorems
100

What is Chain Rule

The derivative of the outside function with the same inside function, multiplied by the derivative of the inside function

100

if the limit as x-->1+ does not equal the limit as x-->1-, does the limit as x-->1 exist?

No, it does not.

100

What is the indefinite integral of 1/u

ln|u|+c

100

What is the derivative of 6x^3

18x^2

100

what does Intermeiate value theorem gurantee

This guarantees that if a function iscontinuous  on [a,b], there must exist a c such that f(c) is in between f(a) and f(b)

200

Find the derivative of f(g(x))

f’(g(x)) · g’(x)

200

What must be true for a function to be continuous at a point.

lim as x--> a- = lim as x-->a+ = f(a)

200

What is the integral of sec(x)

ln|sec(x)+tan(x)|+c

200

what is the derivative of sin(x)+tan(x)

cos(x) + sec(x)^2

200

Can the mean value theorem be used with f(x)=|x| on the interval [-1,1]?

No it cannot, f is not differentiable at x=0

300

find the antiderivative of 4sec(3x)tan(3x)

4/3(sec(3x)) + c

300

what does removable discontinuity mean

the lim as x-->a exists and does not equal f(a)

300

what is the integral of 3x^2 from x=1 to x=4

63

300

what is the derivative of x^2 · ln(x)

2x · ln(x) + x

300

what is the derivative of 2t^2 +5t +7 from t=5 to t=x

2x^2 +5x +7

400

Find the derivative of (5x^2+20x+40)^1/2

(5x+10)/((5x^2+20x+40)^1/2)

400

what is the limit as x--->1 of (x^2 - 1)/(x-1)

2

400

find the definite integral of 5/x  + 9 from 2 to 5

5 ln (5/2) + 27

400

what is the derivative of ln(x^2 +1)

2x/(x^2 + 1)

400

F(x) = the integral from 2 to x^2 of cost(t^2). What is F'(t)

2xcost(x^4)

500

find the derivative of: e^(sin(3x^2))

e^(sin(3x^2))·cos(3x^2)·6x

500

lim as x--> 0 of (e^x -1)/x

Use L'Hopitals rule, 1

500

What is the integral of 1/(4+x^2) 

1/2 · arctan(x/2) +c

500

what is the derivative of sin(x)/4x

(x·cos(x) - sin(x))/4x^2

500

F(x) = x^3 - 2x - 1 is continuous on the interval [1,2]. Is there a c in (1,2) such that f(c) =0

Because f is continous on [1,2] and f(1)=-2 and f(2) = 3, IVT gurantees that there exists a c on (1,2) such that f(c) =0

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