Question

plz help asap very urgent : Find dy/dx , given y =integral( top is exponential^x , bottom is 1 ) (1+surt x )^5 dx

Answer 1

\[ \dfrac{d}{dx}\int \limits _1^{e^x} (1 + \sqrt{x})^5 dx = (1 + \sqrt{e^x})^5 (e^x)'\]

Answer 2

yes correct
explain plz

Answer 3

use FTC + chainrule

Answer 4

sorry , wat is ftc

Answer 5

hello ? explain plz....

Answer 6

http://classconnection.s3.amazonaws.com/33333/flashcards/644875/jpg/ftc1-leibniz-a.jpg

Answer 7

derivative of an integral gives you the function back

Answer 8

\(\dfrac{d}{dx} \int f = f\)

Answer 9

ok , so that means there is no long calculation to do , just sub in the top ?

Answer 10

Yep !

Answer 11

but why is there another ((ex) at the back ?

Answer 12

say \(F(x) = \int (1+\sqrt{x})^5 ~dx\)

Answer 13

ok, continue

Answer 14

the, \( \int \limits_a^t (1+\sqrt{x})^5 ~dx = F(t) - F(a)\)

Answer 15

now take the derivative

Answer 16

oyeah cahin rule u just mentioned , I forgot , haha , thx a thousand :D , actually Im a small mentor in school , then one student came and ask this , now I can help him back :D thanks

Answer 17

\( \dfrac{d}{dt}\int \limits_a^t (1+\sqrt{x})^5 ~dx = \dfrac{d}{dt} \left(F(t) - F(a)\right) = F'(t) - 0 = F'(t)\)

Answer 18

oh ok :)
since "t" itself is another function, we need to use chain rule

Answer 19

ya thx I understood :D

Answer 20

np :)

Answer 21

ftc = fundamental theorem of calculus

Answer 22

^^ and btw the `variable in d/dx` and the` upper bound` need to match " :

\[\dfrac{d}{d \color{red}{x}} \int \limits_a^{\color{red}{x}} f(t) ~ dt = f(x)\]

Answer 23

if you have a function instead of just "x" :

\[\dfrac{d}{d \color{red}{x}} \int \limits_a^{\color{red}{g(x)}} f(t) ~ dt = f(\color{red}{g(x)}) (\color{red}{g(x)} )'\]

Answer 24

u mean it must be x if its dydx or j if its dydj ?

Answer 25

exactly ! it has to match, other wise you wud get 0 :

d/dx (f(t)) = 0

Answer 26

huh interesting. Thank you!!! this clears up so much in my multi-var class right now

Answer 27

lol :P

Answer 28

:) to make it clear :

\[\dfrac{d}{d \color{red}{x}} \int \limits_a^{\color{red}{q}} f(t) ~ dt = 0 \]

Answer 29

ok im stuck help

Answer 30

ganeshie u here ??

Answer 31

im here, is that attachment a different problem ?

Answer 32

no its the same , thats my calculation and im stuck

Answer 33

we already finished hte original problem, right ?

Answer 34

its exactly that question I ask , the attachment is my work , I follow wat u say and do now Im stuck

Answer 35

\[\dfrac{d}{dx}\int \limits _1^{e^x} (1 + \sqrt{x})^5 dx = (1 + \sqrt{e^x})^5 (e^x)' = e^x(1 + \sqrt{e^x})^5 \]

we're done.

Answer 36

u mean I did wrongly ?? did u check my attachment ?

Answer 37

no, i mean we're done wid the problem. we dont need to do any integration... cuz the integral and derivative eat eachother out by fundamental theorem of calculus

Answer 38

u chain rule this ? --->> (1+ex−−√)5

Answer 39

\(e^x(1+\sqrt{e^x})^5\) is the final answer

Answer 40

I know thats the ans

Answer 41

hmm

Answer 42

sorry im still a bit confuse about the extra e^x, u say chain rule , but with wat ?

Answer 43

chain rule for the top bound

Answer 44

lets go thru it step by step again

Answer 45

do u have a paper can u do the chain rule clearly ??

Answer 46

you want to work below :

\[

\dfrac{d}{dx}\int \limits _1^{e^x} (1 + \sqrt{x})^5 dx

\]

Answer 47

to make it less confusing, lets change the dummy variable under integration :

\[

\dfrac{d}{dx}\int \limits _1^{e^x} (1 + \sqrt{t})^5 dt

\]

Answer 48

fine wid this ?

Answer 49

ok

Answer 50

good,

say \(F(t) = \int (1+ \sqrt{t})^5 dt ~~~~~\color{red}{*}\)

Answer 51

\(\implies \)

\[

\int \limits _1^{e^x} (1 + \sqrt{t})^5 dt = F(e^x) - F(1)

\]

Answer 52

ok

Answer 53

still fine ? i have just taken the bounds...

Answer 54

ya

Answer 55

next differentiate both sides with respect to x

Answer 56

\[

\int \limits _1^{e^x} (1 + \sqrt{t})^5 dt = F(e^x) - F(1)

\]
\(\implies \)
\[

\dfrac{d}{dx}\int \limits _1^{e^x} (1 + \sqrt{t})^5 dt = \dfrac{d}{dx} \left[F(e^x) - F(1)
\right]
\]

Answer 57

still wid me ?

Answer 58

ya

Answer 59

is F (t) this ?

Answer 60

yes, but we dont need to work any integral here.

Answer 61

ok ( they cancel eaxh other right ? ) , continue

Answer 62

integral just disappears by Fundamental thm of calculus,
lets do the next step, and it wil be obvious :)

Answer 63

ok , Im curious and excited :)

Answer 64

\[\dfrac{d}{dx}\int \limits _1^{e^x} (1 + \sqrt{t})^5 dt = \dfrac{d}{dx} \left[F(e^x) - F(1)
\right] = \dfrac{d}{dx} \left[F(e^x)
\right] - \dfrac{d}{dx} \left[ F(1)
\right] \\ = \dfrac{d}{dx} \left[F(e^x)
\right] - 0 = \dfrac{d}{dx} \left[F(e^x)
\right] \]

Answer 65

still fine ?

Answer 66

giv me a moment

Answer 67

ok, continue

Answer 68

take ur time :) btw, earlier we defined F(t) as :

\(F(t) = \int (1+ \sqrt{t})^5 dt ~~~~~\color{red}{*} \)

Answer 69

By FTC, taking the derivative gives u the funciton back :

\(F(t) = \int (1+ \sqrt{t})^5 dt ~~~~~\color{red}{*} \)

\(\implies \dfrac{d}{dt}F(t) = (1+\sqrt{t})^5\)

Answer 70

ok

Answer 71

\[
\dfrac{d}{dx}\int \limits _1^{e^x} (1 + \sqrt{t})^5 dt = \dfrac{d}{dx} \left[F(e^x) - F(1)
\right] = \dfrac{d}{dx} \left[F(e^x)
\right] - \dfrac{d}{dx} \left[ F(1)
\right] \\ = \dfrac{d}{dx} \left[F(e^x)
\right] - 0 = \dfrac{d}{dx} \left[F(e^x)
\right]

\]

Answer 72

use the previous result( \(\color{red}{*}\)) to evaluate above ^

Answer 73

ok, but there is no second e^x?

Answer 74

good question, say \(e^x = u\)

Answer 75

By chain rule :

\[
\dfrac{d}{dx} \left[F(e^x)
\right] = \dfrac{d}{dx} \left[F(u)
\right] * \dfrac{d}{dx} [u]

\]

Answer 76

\[= (1+\sqrt{u})^5 \dfrac{d}{dx} [u]\]

Answer 77

plugin the value of \(u\)

Answer 78

a moment plz.... digesting

Answer 79

k

Answer 80

if u chain rule , isnt the power 5 suppose to bring to the front and the original minus by 1 ?

Answer 81

nope,


\[F(t) = \int (1+ \sqrt{t})^5 dt ~~~~~\color{red}{*} \]
\[ \implies \dfrac{d}{dt}F(t) = (1+\sqrt{t})^5 \]

Answer 82

^^thats directly by Fundamental theorem of caluclus,
you okay wid that right ?

Answer 83

im not sure about the chan rule , cuz mine is completely different , I let u see , wait I snap photo

Answer 84

are you fine wid Fundamental theorem of calculus above ?

Answer 85

lets finish this, before moving to seeing ur work

Answer 86

too many things = too many confusions lol

Answer 87

wait , u just have a look at this first

Answer 88

looks u have attached the wrong pic, check once

Answer 89

yes its this , look at the chain rule

Answer 90

no.8

Answer 91

yes

Answer 92

why it brings the power to the front ?

Answer 93

thats the chain rule for taking derivative of a specific function of type : \([f(x) ]^n\)

Answer 94

ok, which no. should I be looking at ?

Answer 95

this is not in the table ?

Answer 96

a more general chain rule :

\[y = f(g(x)) \implies \dfrac{dy}{dx} = f'(g(x)) \times g'(x)\]