Question

the derivative of square root of t is 2t?

Answer 1

no it's 1/2t

Answer 2

because square root is the same as t^1/2 and you bring that to the front in derivatives.

Answer 3

\[\frac{1}{2\sqrt{t}}\]

Answer 4

oh yeah hehe brain fart

Answer 5

can anyone help me in product rule?

Answer 6

on*

Answer 7

Yeah i can.

Answer 8

okay so I have \[2x ^{3}e ^{7x}\] and i have to use product rule. So almost at the final stage I got (2x^3)(7e^(7x))+(e^7x)(6x^2)...but I'm a bit confused on how to multiply them and simplify, especially witth the exponents

Answer 9

Ok i will start from the beginning so i dont get confused lol.

Answer 10

\[(6x^2)(e^{7x}) + (2x^{3})(e^{7x})\]

Answer 11

yes thats what i got, now how to simplify

Answer 12

I might of taken the derivative of \[e^{7x}\] wrong.. let me check before i move on.

Answer 13

Yeah i did. Let me correct it.

Answer 14

oh yeah u did, it was suppose to be 7 in front

Answer 15

\[(6x^{2})(e^{7x}) + (3x^{2})(7e^{7x})\]

Answer 16

Ok we can factor out a e^7x

Answer 17

okay wait, its suppose to be uv'+vu' form...

Answer 18

\[e^{7x}[6x^{2} + 3x^{2}*7]\]

Answer 19

The order doesnt matter in the product rule.

Answer 20

okay but shouldnt the first parentheses ont he second set be \[2x ^{?}\] not 3x^2

Answer 21

i mean 3x^3

Answer 22

ahh i bumped the three and i ment the 2.. Lets see if i can get it right this time >.< lol

Answer 23

\[(6x^2)(e^{7x})+(2x^3)(7e^{7x})\]

Answer 24

Ok did i get it right this time?? (i hope lol )

Answer 25

Factor out e^7x \[e^{7x}[6x^2 + 14x^3]\] and you can also factor out a 2x^2 \[e^{7x}2x^2[3 + 7x]\] And that is the answer!

Answer 26

http://www.wolframalpha.com/input/?i=%282x%5E3%29%28e%5E%287x%29%29&t=mfftb01
This confirms i didnt screw it up this time lol

Answer 27

okay thanks! can you help me with one more?

Answer 28

Sure

Answer 29

f(x)=\[-x ^{2}e ^{5\sin(x)}\]

Answer 30

Ok \[(-2x)(e^{5\sin(x)}) + (-x^2)(5e^{5\sin(x)}\cos(x))\]

Answer 31

Factor out -xe^5sin(x)\[-e^{5\sin(x)}x[2 + 5xcos(x)]\]

Answer 32

okay for the second set, it couldnt have been written like 3cos(x)e^(5sinx)?

Answer 33

or does the cos have to be beside 5e adn its exponent

Answer 34

by 3 i meatnt 5

Answer 35

it would be \[\cos(x)5e^{5\sin(x)}\] if you wanted the cosine in the front. The way i have it written is correct though and its just force of habbit when i use the chain rule i write the derivative of the inner function last.

Answer 36

okay, well we havent been taught the chain rule yet. we are next week.

Answer 37

Wow. You need the chain rule to solve the last one lol.. Are you working ahead in class or something or is your teacher that mean lol :P

Answer 38

perhaps she didnt realize it required the chain rule. I don't know. She's not mean, but she expects us to know stuff :O ...but y exactly do we need the chain rule for it?

Answer 39

Because the e^5sin(x) in a composite function. And to find the derivative of a composite function you need the Chain Rule.

Answer 40

Its pretty simple to learn, i can teach it to you real quick if you want.

Answer 41

oh actually we skipped to chapter 5 for those involiving e....natural exponential functions and how to find their derivatives....(is that the same or different?)

Answer 42

Same or different as what??

Answer 43

the same as finding the derivative with the chain rule.. because this is the form to gett he derivatives involving e f'(u)=(e^u)du

Answer 44

Yeah its kinda like the chain rule.

Answer 45

okay. so what is the chain rule? Can you teach me? I'll gladly appreciate it. That way i can be ahead in knowing.

Answer 46

Ok.. Chain rule is used to find the derivative of Composite Functions in the form \[f(g(x))\] ( f composed with g(x) ).. Where f(x) is the outer function and g(x) is the inner function.

Answer 47

So say we have the function \[y = (2x + 1)^2\]

Answer 48

The outer function is \[f(x) = x^2\] and the inner function is \[g(x) = 2x + 1\]. F is composed with g(x). Do you understand so far?

Answer 49

yes I do but why is the outer function xsquared?

Answer 50

Its like doing the composition of a function in reverse. You have \[f(g(x)) = (2x + 1)^2\] What is f(x) and what is g(x).. f(x) is the ^2 and g(x) is 2x + 1

Answer 51

ok got it

Answer 52

i just use x because it is the most common variable used. It can be y^2 or z^2 ect...

Answer 53

Ok.. before we move on, just to make sure you understand \[f(g(x)) = \sqrt{2x^2 + 3}\] What is f(x) (outer function) and what is g(x) (inner function)??

Answer 54

inner function is 2x^2+3 and outer function is x^1/2?

Answer 55

Good!

Answer 56

Ok, if we have a composite function in the form f(g(x)) then to find the derivative you do the following. \[y \prime = f \prime(g(x))* g \prime (x)\]

Answer 57

You take the derivative of the outer function and leave the inner function alone then multiply it by the derivative of the inner function.

Answer 58

So going back to the example problem \[(2x+1)^2\] The outer function is \[f(x) = x^2\] and the inner function is \[g(x) = 2x + 1\]

Answer 59

The chain rule first says to find the derivative of the outer function. So the derivative of \[x^2\] is \[2x\] and to put g(x) where x is so it will look like \[2(2x+1)\] then multiply it by the derivative of 2x + 1 which is 2. So it will look like this \[2(2x+1) * 2\]

Answer 60

Make sense??

Answer 61

yes yes

Answer 62

Then after you simplify it you will end up with \[8x + 4\]

Answer 63

Its pretty simple once you know how to do it :D

Answer 64

hold on, i believe a question is coming in..

Answer 65

okay got it! This is kind of awkward to ask but do you have a facebook page?

Answer 66

You sure you understand? And i do.

Answer 67

Yeah I understand completely what you told me. Do you think you can add me as a friend?

Answer 68

Sure.

Answer 69

okay my email is cald.dany@yahoo.com

Answer 70

Do you have pickachu as a profile pic??

Answer 71

hah yeah

Answer 72

thats me