Maths1210
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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.
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Pangol
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(Original post by 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.
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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afx42
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If you take out the log to get e^x = 2y+3 then rearrange everything it should be relatively straightforward
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the bear
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find dx/dy then flip it over
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Maths1210
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(Original post by 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?
I dont know what you mean by writing it in terms of x.
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mnot
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(Original post by Maths1210)
I dont know what you mean by writing it in terms of x.
So rearrange into the form y = f(x) (raise both sides to e^function).
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Maths1210
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(Original post by mnot)
So rearrange into the form y = f(x) (raise both sides to e^function).
i have no idea how to do this
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Maths1210
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(Original post by Maths1210)
i have no idea how to do this
wait is it from x=ln(2y+3) ?
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mnot
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(Original post by Maths1210)
i have no idea how to do this
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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Maths1210
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(Original post by 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...
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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RDKGames
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(Original post by 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.
The RHS is simply 2y+3. Do you understand why ?
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mnot
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(Original post by 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.
this will simplify to e^x = 2y + 3

e^x and ln(x) are inverses, its important you go and learn whats happening here.
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Maths1210
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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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Maths1210
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(Original post by 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
I have 1/2e^(x) but still dont fully understand how I got here
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RDKGames
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(Original post by 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
You seem very confused... so I'll just put this to bed because it really shouldn't be that hard.

At e^x = 2y+3 you differentiate both sides with respect to x and be left with e^x = 2\dfrac{dy}{dx}.
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Maths1210
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I just don't get how you got 2 dy/dx on the RHS
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RDKGames
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(Original post by Maths1210)
I just don't get how you got 2 dy/dx on the RHS
The +3 is a constant, so obviously this disappears when you differentiate it.

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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Maths1210
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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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RDKGames
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(Original post by 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
I suggest you cover the basics of differentiation and notation.

We would only get 2 on its own if it was 2x+3 because we are differentiating with respect to x, but we have 2y+3 instead... and y is a function of x.
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dan_klim
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(Original post by 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
That is the rule of implicit differentiation. Do you have a maths textbook?
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