Question

can i have help with this question pretty pleeease!!

Answer 1

Where is it @

Answer 2

question 4 part (ii)

Answer 3

@campbell_st pleaseeee

Answer 4

my guess its the one with 2 black stars

Answer 5

so you found part 1
\[x=\frac{\ln(y)}{\ln(a)} \\ \frac{dx}{dy}=\frac{1}{\ln(a)} \frac{d}{dy}(\ln(y))\]

Answer 6

your only objective is to really differentiate ln(y) w.r.t y

Answer 7

if its the one with the black stars firstly expand those exp

Answer 8

then separately integrate eachone of them

Answer 9

so is it the question

\[\int\limits(e^x - e^{3x})(e^x + e^{3x}) dx\]

Answer 10

these are my attemps, and it's question 4 part (ii) i'm looking at

Answer 11

no question 4 part (ii)

Answer 12

ok... so multiply it out, its the difference of 2 squares

\[\int\limits e^{2x} - e^{6x}~ dx \]

does that make it easier

Answer 13

@study17 you are aware ln(a) is a constant
and therefore 1/ln(a) is also a constant?

Answer 14

\[x=\frac{\ln(y)}{\ln(a)} \\ x=\frac{1}{\ln(a)} \cdot \ln(y) \\ \text{ find the derivative of both sides w.r.t } y\\ \frac{d}{dy}(x)=\frac{d}{dy}( \frac{1}{\ln(a)} \cdot \ln(y)) \\ \frac{dx}{dy}=\frac{1}{\ln(a)} \frac{d}{dy}(\ln(y) ) \text{ since } \frac{1}{\ln(a) } \text{ is a constant } \]

Answer 15

as i said your only objective is to differentiate ln(y) w.r.t. y
do you know how to do that

Answer 16

ok... so you need part (1) and rewriting it

\[x = \frac{1}{\ln(a)} \times \ln(y)\]

you are asked to find the derivative of x with respect to y.

\[\frac{1}{\ln(a)}\] is a constant

so you only need to look at differentiating

\[\ln(y) \] with respect to y. and then multiply the answer by

\[\frac{1}{\ln(a)}\]

does that make sense

Answer 17

wait what? i thought i just had to use the quotient rule, why did i have to rewrite part (i), i don't get that?

Answer 18

you could use the quotient rule but it is a lot of unnecessary work

Answer 19

well the ln(a) part is a constant

s an example x/2 would have 1/2 as a constant

Answer 20

\[\frac{dx}{dy}=\frac{\ln(y) \cdot \frac{d}{dy}(\ln(a))-\ln(a) \cdot \frac{d}{dy}(\ln(y))}{(\ln(a))^2} \\ \frac{dx}{dy}=\frac{\ln(y) \cdot 0-\ln(a) \cdot \frac{1}{y}}{\ln(a))^2}\]

Answer 21

why is it a constant? i really don;t understand

Answer 22

this will give you the same thing if you have just used the constant multiple rule in the beginning

Answer 23

well because it says a is a constant

Answer 24

It says "Consider the function y=a^x, where a is some constant"

Answer 25

ok... a is the base with an exponent x..

such as 2 or 3 and the question says so

Answer 26

and I did the quotient rule a little backwards let me fix that

Answer 27

yeah okay i understand that but i still don't know how A being a constant changes how i do the question, i genuinely just thought i had to use the quotient rule

Answer 28

that is why you don't need the quotient rule

Answer 29

\[\frac{dx}{dy}=\frac{-\ln(y) \cdot \frac{d}{dy}(\ln(a))+\ln(a) \cdot \frac{d}{dy}(\ln(y))}{(\ln(a))^2} \\ \frac{dx}{dy}=\frac{-\ln(y) \cdot 0+\ln(a) \cdot \frac{1}{y}}{\ln(a))^2} \]

Answer 30

well by rewriting it... you get you could use the product rule

and the derivative of a constant is 0.

x = 1/ln(a) * ln(y)

x' = 0 * ln(y) + 1/ln(a)* 1/y

so you get the derivative with respect to y

Answer 31

@study17 you can use the quotient rule...but there is a lot of odd things in your use of the rule:
I'm very unsure how you got your very first line so let's start there:
\[\frac{dx}{dy}=\frac{d}{dy}(\frac{\ln(y)}{\ln(a)}) \\ =\frac{\ln(a) \cdot \frac{d}{dy} (\ln(y))- \ln(y) \cdot \frac{d}{dy} (\ln(a))}{(\ln(a))^2}\]
do you agree with this so far
I'm using the quotient rule

Answer 32

from your notes it looks like you know the derivative of ln(y) w.r.t. to y is 1/y
no derivative of ln(3) w.r.t y is 0 right?
so if a is a constant then so is ln(a)
and the derivative of ln(a) w.r.t y is also 0 right?

Answer 33

\[\frac{dx}{dy}=\frac{d}{dy}(\frac{\ln(y)}{\ln(a)}) \\ =\frac{\ln(a) \cdot \frac{d}{dy} (\ln(y))- \ln(y) \cdot \frac{d}{dy} (\ln(a))}{(\ln(a))^2} =\frac{\ln(a) \frac{1}{y}-\ln(y) \cdot 0}{(\ln(a))^2} \\ =\frac{\ln(a) \frac{1}{y}}{(\ln(a))^2}\]
you can simplify this more

Answer 34

\[=\frac{\ln(a)}{(\ln(a))^2} \frac{1}{y} \\ =\frac{1}{\ln(a)} \frac{1}{y}\]

Answer 35

which gives us the very same thing if we would have used the constant multiple rule or even the product rule as @campbell_st mentioned

Answer 36

ok i don't want to use the quotient rule then if that's the case, what's the easier way? and @campbell_st what did you do? i'm kinda confused when you write it like that , do you mind writing normal please?

Answer 37

\[x=\frac{\ln(y)}{\ln(a)} \\ x=\frac{1}{\ln(a)} \cdot \ln(y) \\ \text{ find the derivative of both sides w.r.t } y\\ \frac{d}{dy}(x)=\frac{d}{dy}( \frac{1}{\ln(a)} \cdot \ln(y)) \\ \frac{dx}{dy}=\frac{1}{\ln(a)} \frac{d}{dy}(\ln(y) ) \text{ since } \frac{1}{\ln(a) } \text{ is a constant } \]
This is what I wrote above

Answer 38

this is the easier way
just using the constant multiple rule

Answer 39

you only need to differentiate ln(y) w.r.t y now

Answer 40

so the anser is 1/lna(y) ?

Answer 41

yes I think you mean this right:\
[\frac{dx}{dy}=\frac{1}{\ln(a) \cdot y}\]

Answer 42

\[\frac{dx}{dy}=\frac{1}{\ln(a) \cdot y}\]

Answer 43

yup, i still don't know why i had to rewrite it as 1/lna * lny and why i only had to differentiate lny, like i wouldnt know that i had to that if it was an exam ..

Answer 44

do you know what is meant by a constant?

Answer 45

like for example so you are saying you wouldn't know how to differentiate (w.r.t y):
\[x=\frac{\ln(y)}{\ln(2)}\]

Answer 46

yeah a number?

Answer 47

that a is a number just like 2 or 3 or 6 or 1041.5252

Answer 48

we are told this in the problem

Answer 49

yeah i know what a constant is, like would i not have to use the quotient rule if i was asked for the dy/dx of lny/ln2 ?

Answer 50

a constant number is a number that always is and never changes
so the change is 0 therefore the derivative is 0

Answer 51

you could

Answer 52

i used the quotient rule above for any number a

Answer 53

a>0 at least

Answer 54

since ln(a) only exists for a>0

Answer 55

\[x=\frac{\ln(y)}{\ln(2)} \\ \frac{dx}{dy}=\frac{\ln(2) \cdot \frac{d}{dy}\ln(y)-\ln(y) \cdot \frac{d}{dy} \ln(2)}{(\ln(2))^2} \\ \frac{dx}{dy}=\frac{\ln(2) \cdot \frac{1}{y}-\ln(y) \cdot 0}{(\ln(2))^2}=\frac{\ln(2) \cdot \frac{1}{y}}{(\ln(2))^2} =\frac{\ln(2)}{(\ln(2))^2} \cdot \frac{1}{y}=\frac{1}{\ln(2)} \cdot \frac{1}{y} =\frac{1}{y \ln(2)}\]

Answer 56

but using the constant multiple rule is faster

Answer 57

\[x=\frac{\ln(y)}{\ln(2)}=\frac{1}{\ln(2)} \cdot \ln(y) \\ \frac{dx}{dy} =\frac{1}{\ln(2)} \frac{d}{dy}(\ln(y))=\frac{1}{\ln(2)} \cdot \frac{1}{y} =\frac{1}{y \ln(2)}\]

Answer 58

if you see a number called a constant or if it is used in the context as a constant
then treat it like a constant (treat it like you would treat 2 or 5 or ln(5))

Answer 59

oh okay, do you know how to do part (iii) ?

Answer 60

we are given that dy/dx=1/what we just found

Answer 61

\[\frac{dy}{dx}=\frac{1}{\frac{1}{y \ln(a)}}\]

Answer 62

what is the reciprocal of the 1/(y ln(a))?

Answer 63

example:
the reciprocal of 1/m is m

Answer 64

these are my attempts

Answer 65

\[\frac{dy}{dx}=\frac{1}{\frac{1}{y \ln(a)}} =y \ln(a)\]
yes the reciprocal is (1/y ln(a)) is y ln(a)
after that it looks like you are doing unneeded work because you are basically done with part 3

Answer 66

just replace y with a^x

Answer 67

and you can do this since y=a^x

Answer 68

are you there?
are you confused about something?
Like we already found dy/dx
you don't need to differentiate anything in step 3

Answer 69

we found dy/dx by finding 1/(dx/dy)

Answer 70

dy/dx=y ln(a)

we know from the very get go that y=a^x
so we can rewrite this as
dy/dx=a^x ln(a)

Answer 71

@freckles oh ok thanks so much!