# Integrations and Derivations

1. Nov 3, 2007

### Aftermarth

Ok im in my final study for my last exams... and i just want to check with all you guys that i have my calculus right :)

Derviations:
1. Lower the power by one, and times by the new power out in front of the term. Coefficinets drop off, and terms with just x become a new coefficient in the f ' (x) form.
2. e ^ x dervies to:
(f '(x)) e ^ x
3. ln (f(x)) dervies to
[ f '(x)] / [f(x)]
4. Chain rule
( 3x + 4)^5 becomes:
4(3x + 4)^4 x 3
=12(3x + 4)^4

5. Product Rule:
vu' + uv' where v and u are both functions multiplied together

6. Quotient Rule:
(vu' - uv') / (2v)

f ''(x) is simply the derivative of the derivative

Integration:
1. Raise the power by one and mutliply by 1/new power. Add the constant term (c) and any coefficients now add the term x
eg. x^3 + 2x^2 + 4
integrates to:
(1/4)x^4 + (2/3)x^3 + 4x + c

2. f '(x) / f(x) integrates to ln( f(x)) + c
3. e ^ f(x) integrates to (1/ f '(x)) e^(f(x)) +c
4. There are no chain, product or quotient rule to integration
5. Always add the constant term c, and use a given point to work it out :)

i hope u guys can identify any probelms in my working if there are any :)
Aftermarth

2. Nov 3, 2007

### rocomath

4 is wrong

your power is 5, not 4

6 is wrong

the denominator is squared not multiplied by 2

3. Nov 3, 2007

### Aftermarth

u mean 4 in derviations?
doesn't the power always get lowered in deriving things?

and six. yes my bad thank you :)

4. Nov 3, 2007

### rocomath

sorry i wasn't clear enough

your power was 5, so you bring it down

you brought down a 4, the reduction process was correct tho

5. Nov 3, 2007

### Aftermarth

ok so.
4. Chain rule
( 3x + 4)^5 becomes:
4(3x + 4)^4 x 3
=12(3x + 4)^4

where the coefficient i brang down is 4 (the new power) is that wrong?
should it be that i bring down 5 instead (the old power) like so:

( 3x + 4)^5 becomes:
5(3x + 4)^4 x 3
=15(3x + 4)^4

6. Nov 3, 2007

### rocomath

yes, that is correct

$$y=(3x+4)^{5}$$

$$y'=5(3x+4)^{5-1}3$$

$$y'=15(3x+4)^{4}$$

7. Nov 3, 2007

### Aftermarth

ok thank you :)

the ln and e integrations and derviations are ok... they are the ones im not so good at

8. Nov 3, 2007

### rocomath

just remember, the derivative of e is itself times the derivative of it's power

$$y=e^{x}$$

$$y'=e^{x}$$

now with a higher power

$$y=e^{2x^{2}}$$

$$y'=e^{2x^{2}}\times 4x$$

if your power happens to be a product, then you do the product rule, etc.

9. Nov 3, 2007

### rocomath

now for ln

$$y=\ln{x}$$

$$y'=\frac{1}{x}$$

now a step further

$$y=\ln{2x^{3}+4x$$

$$y'=\frac{6x+4}{2x^{3}+4x}$$

10. Nov 3, 2007

### Aftermarth

okies thats fine :)
and integration?

11. Nov 3, 2007

### rocomath

you have the concept of integration down, but what if your power was a fraction?

$$\int\sqrt{x}dx$$

???

12. Nov 3, 2007

### Aftermarth

therefore:
integration of x^(1/2)
raise power by one...
x^(3/2)
divide by the new power

= 1/(3/2) x ^(3/2)
=(2/3) x ^(3/2)

13. Nov 3, 2007

### rocomath

yep, it's basically the reciprocal of the power

you pretty much have it down, just be cool b4 the test. you'll kick it's ass, no worries :-]

14. Nov 3, 2007

### Aftermarth

sweet thanks :) ur help is much appreciated :D

15. Nov 3, 2007

### rocomath

anytime

btw, you forgot +C ... :p

16. Nov 3, 2007

### Aftermarth

i always forget that :(. but im sure i will remember unless they give me a point on the curve and tell us to work out c, and they generally do that alot cause i have done heaps of past papers.

just question regarding the graphs of derivatives etc.

When f '(x) = 0 is a turning point for f(x)
When f '(x) is positive, the gradient of f(x) is also positive and vice-versa

when f ''(x) = 0 is a point of inflection for f(x)
when f "(x) is positive means concave up for f(x) and vice - versa

is that right?

17. Nov 20, 2007

### Aftermarth

okies final exam done now

there was much calculus in it and im glad i did my study :)
thank you i did really well :D