Question

i think i figured this equation out but i need someone to check it.

Answer 1

\[xe ^{y} + 1 = xy\]

Answer 2

\[\frac{ xe ^{y} + 1 }{ x} = y\]

Answer 3

\[xe ^{y} + 1 \times (x ^{-1})\]

Answer 4

\[(xe ^{y} + 1) (-1) + (x^{-1})(xe^{y} + 0)\]

Answer 5

\[-(xe^{y} + 1 ) + (x^{-1})(xe^{y})\]

Answer 6

i dont understand wht u need to do on that equation?

Answer 7

finding the derivative.

Answer 8

\[(-xe^{y} -1) + (x^{-1})(xe^{y} + 0)\]

Answer 9

well i really dont understood, sorry, do u know implicit diferentiation?

Answer 10

we've done it in class but i don't remember how to do it.

Answer 11

i think that the way to do this is by implicit diferentiation

Answer 12

i dont understodd the method that u used to solve this

Answer 13

which metod it was

Answer 14

the chain rule to differentiate.

Answer 15

i really dont understad what u did
if u could explaint me each step
or we can try with implicit diferentiation

Answer 16

try the implicit differentiation

Answer 17

we have this
xe^y +1=xy

Answer 18

okay

Answer 19

u remember the product rule right?

Answer 20

first times the dx/dy of the second plus second times the dx/dy of the first?

Answer 21

yes lets work with just this part
xe^y
i meant the firts term

Answer 22

u got it?

Answer 23

which is ...\[x(e^{y}) + e^{y} (1) + 0\]

Answer 24

yes but u must get the derivative of (y) too

Answer 25

and put it where?

Answer 26

x(e^y)(dy/dx)+ey(1)+0

Answer 27

x(e^y)(dy/dx)+e^y(1)+0

Answer 28

did u get it?

Answer 29

yes

Answer 30

remember the original equation equation

Answer 31

xe^y +1=xy

Answer 32

now lets work with the (xy)

Answer 33

i meant the right part of the equation

Answer 34

ok

Answer 35

are u following me?

Answer 36

ok could u apply the product rule on
xy

Answer 37

yes

Answer 38

x(dy) + y(dx)

Answer 39

i m not sure about that

Answer 40

xy´ + yx´
that is what we r trying to find

Answer 41

okay

Answer 42

is that right?

Answer 43

ok but y´ = dy/dx
and x´=1

Answer 44

yes

Answer 45

do u got that?

Answer 46

how come x isn't dx/dy

Answer 47

sorry x'

Answer 48

we r trying to find the derivative with respect to the variable x

Answer 49

so for y we have y´ = dy/dx

Answer 50

and for x we have x´ = dx/dx = 1

Answer 51

ohh okay

Answer 52

so lets write the derivative of xy

Answer 53

x + y(dy/dx)

Answer 54

well no

Answer 55

again with the product rule we have

Answer 56

x´y + xy´

Answer 57

x´ = 1
y´ =dy/dx

Answer 58

y + x(dy/dx)

Answer 59

yes now lets put all the parts together in the original equation
xe^y +1=xy

Answer 60

could u do that

Answer 61

\[x(e^{y})\frac{ dy }{ dx} + e^{y} = y + x(\frac{ dy }{ dx })\]

Answer 62

that is right
now we must solve for dy/dx

Answer 63

do u know how to do that?

Answer 64

i can trytry

Answer 65

ok
will be wathching

Answer 66

jajaaj bad english

Answer 67

\[\frac{ e^{y}-y }{ x-xe^{y}}=\frac{ dy }{ dx}\]

Answer 68

yes

Answer 69

so it is done

Answer 70

could you you help me on one more?

Answer 71

ok i will try

Answer 72

\[\int\limits \frac{ e^\sqrt{x} }{ \sqrt{x}} dx\]

Answer 73

have u tried u-substitution?

Answer 74

i thought so,but im not sure if i use square root of x

Answer 75

yes\[u=\sqrt{x}\]

Answer 76

now lets find du/dx

Answer 77

-1/2x^1/2

Answer 78

yes

Answer 79

so we multiply by 1/2

Answer 80

or e?

Answer 81

e?

Answer 82

im stuck right theere

Answer 83

lets write it with the diferentials

Answer 84

what you mean?

Answer 85

du= \[\frac{ 1 }{ 2\sqrt{x} }\]dx

Answer 86

i don't understand

Answer 87

\[du = \frac{ 1 }{ 2\sqrt{x} }dx\]

Answer 88

do u get it now?

Answer 89

so its not ..\[-\frac{ 1 }{ 2 }x^{\frac{ 1 }{ 2 }}\]dx

Answer 90

no

Answer 91

remember the power rule for derivatives

Answer 92

okay

Answer 93

so u know the reason?

Answer 94

not sure,but your du = my du i just got rid of the radical sign and the fraction

Answer 95

wait .. aren't i suppose to subtract one fro the exponent when i bring it to the front?

Answer 96

yes

Answer 97

so .. what did i do wrong?

Answer 98

the power rule is
\[d x^{n}/dx = nx ^{n-1}\]