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How would you find dy/dx when x=ln (2y+3)?

Before I did it using the chain rule but got the answer (2y+3)/2 but the answer says 1/2e^(x). I'm not sure how you get to there.

Before I did it using the chain rule but got the answer (2y+3)/2 but the answer says 1/2e^(x). I'm not sure how you get to there.

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#2

(Original post by

How would you find dy/dx when x=ln (2y+3)?

Before I did it using the chain rule but got the answer (2y+3)/2 but the answer says 1/2e^(x). I'm not sure how you get to there.

**Maths1210**)How would you find dy/dx when x=ln (2y+3)?

Before I did it using the chain rule but got the answer (2y+3)/2 but the answer says 1/2e^(x). I'm not sure how you get to there.

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#3

If you take out the log to get e^x = 2y+3 then rearrange everything it should be relatively straightforward

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(Original post by

What you have done is correct. They are expecting an answer in terms of x, not y. If x = ln(2y + 3), can you write 2y + 3 in terms of x?

**Pangol**)What you have done is correct. They are expecting an answer in terms of x, not y. If x = ln(2y + 3), can you write 2y + 3 in terms of x?

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#6

(Original post by

I dont know what you mean by writing it in terms of x.

**Maths1210**)I dont know what you mean by writing it in terms of x.

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(Original post by

So rearrange into the form y = f(x) (raise both sides to e^function).

**mnot**)So rearrange into the form y = f(x) (raise both sides to e^function).

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(Original post by

i have no idea how to do this

**Maths1210**)i have no idea how to do this

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#9

(Original post by

i have no idea how to do this

**Maths1210**)i have no idea how to do this

y = ln(x) ,

raise both sides to e^(f)

you get: e^y = e^(ln(x))

we can then simplify the right hand side so:

e^y = x ; (you see what happens when e^ln or ln(e)...)

now you take this strategy and apply it to this question you can get it into the form y = Ae^(x) + B which will simply to the dy/dx solution you want...

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(Original post by

So ignore the question. but if I have the expression:

y = ln(x) ,

raise both sides to e^(f)

you get: e^y = e^(ln(x))

we can then simplify the right hand side so:

e^y = x ; (you see what happens when e^ln or ln(e)...)

now you take this strategy and apply it to this question you can get it into the form y = Ae^(x) + B which will simply to the dy/dx solution you want...

**mnot**)So ignore the question. but if I have the expression:

y = ln(x) ,

raise both sides to e^(f)

you get: e^y = e^(ln(x))

we can then simplify the right hand side so:

e^y = x ; (you see what happens when e^ln or ln(e)...)

now you take this strategy and apply it to this question you can get it into the form y = Ae^(x) + B which will simply to the dy/dx solution you want...

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#11

(Original post by

Still not getting it. so far I have x=ln(2y+3) then e^(x) = e^(ln(2y+3) but now im just confused and have no idea how to simplify.

**Maths1210**)Still not getting it. so far I have x=ln(2y+3) then e^(x) = e^(ln(2y+3) but now im just confused and have no idea how to simplify.

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#12

**Maths1210**)

Still not getting it. so far I have x=ln(2y+3) then e^(x) = e^(ln(2y+3) but now im just confused and have no idea how to simplify.

e^x and ln(x) are inverses, its important you go and learn whats happening here.

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ohhhhhh so if e^(x) = 2y+3 and I already have dy/dx = (2y+3)/2 then I get e^(x) = (2y+3)/2

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(Original post by

ohhhhhh so if e^(x) = 2y+3 and I already have dy/dx = (2y+3)/2 then I get e^(x) = (2y+3)/2

**Maths1210**)ohhhhhh so if e^(x) = 2y+3 and I already have dy/dx = (2y+3)/2 then I get e^(x) = (2y+3)/2

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#15

**Maths1210**)

ohhhhhh so if e^(x) = 2y+3 and I already have dy/dx = (2y+3)/2 then I get e^(x) = (2y+3)/2

At you differentiate both sides with respect to and be left with .

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#17

(Original post by

I just don't get how you got 2 dy/dx on the RHS

**Maths1210**)I just don't get how you got 2 dy/dx on the RHS

Then 2y differentiates to 2*dy/dx much like if you had y = x^2 you would differentiate this to dy/dx = 2x.

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I know how to differentiate I just don't see how you get 2 x dy/dy because surely 2y+3 would just be 2 right? So im just wondering why you have to times by the derivative. also why you have to differentiate again when you've just differentiated

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#19

(Original post by

I know how to differentiate I just don't see how you get 2 x dy/dy because surely 2y+3 would just be 2 right? So im just wondering why you have to times by the derivative. also why you have to differentiate again when you've just differentiated

**Maths1210**)I know how to differentiate I just don't see how you get 2 x dy/dy because surely 2y+3 would just be 2 right? So im just wondering why you have to times by the derivative. also why you have to differentiate again when you've just differentiated

We would only get 2 on its own if it was because we are differentiating with respect to x, but we have instead... and y is a function of x.

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#20

**Maths1210**)

I know how to differentiate I just don't see how you get 2 x dy/dy because surely 2y+3 would just be 2 right? So im just wondering why you have to times by the derivative. also why you have to differentiate again when you've just differentiated

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