Differentiate y = arccosec(x) Watch

Speleo
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#21
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Try integrating 1/(kt) between suitable limits to "prove" that log(x) + log(k) = log(kx) using only the fact that log1 = 0 and that the derivative of logx is 1/x.
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#22
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(Original post by Speleo)
Try integrating 1/(kt) between suitable limits to "prove" that log(x) + log(k) = log(kx) using only the fact that log1 = 0 and that the derivative of logx is 1/x.
Sad as I am, I just did it, but did you create ths problem yourself? If you did, it's a very impressive formulation.
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Speleo
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#23
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Yeah, I was trying to think of a way to apply this idea not using trigonometry.
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Zii
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#24
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#24
(Original post by Speleo)
Try integrating 1/(kt) between suitable limits to "prove" that log(x) + log(k) = log(kx) using only the fact that log1 = 0 and that the derivative of logx is 1/x.
I've done it between 1 and x, and have gotten ln(kx) - ln(k) and presumably I want to show that this is equal to ln(x). How do I do this?

Edit: I see how to do it now!!! Also yeah I spotted the mistake!
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Speleo
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#25
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#25
There are two ways to think about integrating 1/(kt). You have essentially done the substitution u = kt, but there is another way, how would you integrate 1/ksinx?

(your answer is slightly wrong by the way...)
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Zii
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#26
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(Original post by Speleo)
Try integrating 1/(kt) between suitable limits to "prove" that log(x) + log(k) = log(kx) using only the fact that log1 = 0 and that the derivative of logx is 1/x.
\displaystyle\int^x_1 \frac{1}{kt} \, \mathrm{d}t = \frac{1}{k}(\mathrm{ln}kx - \mathrm{ln}k)

\displaystyle\int^x_1 \frac{1}{kt} \, \mathrm{d}t = \frac{1}{k}\displaystyle\int^x_1 \frac{1}{t} \, \mathrm{d}t = \frac{1}{k}(\mathrm{ln}x - \mathrm{ln}1) = \frac{1}{k}\mathrm{ln}x

\frac{1}{k}(\mathrm{ln}kx - \mathrm{ln}k) = \frac{1}{k}\mathrm{ln}x

\mathrm{ln}kx - \mathrm{ln}k = \mathrm{ln}x

\mathrm{ln}kx = \mathrm{ln}k + \mathrm{ln}x

Q.E.D.
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Speleo
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#27
Using nothing more complicated than the chain rule, prove that the derivative of gol(x), the inverse function of log(x) is gol(x).

i.e. you can use the facts that d/dx log(x) = 1/x, log(1) = 0 (superfluous) and that log[gol(x)] = gol[log(x)] = x

EDIT: also that d/dx x = 1 etc.
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Cities
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(Original post by Speleo)
Using nothing more complicated than the chain rule, prove that the derivative of gol(x), the inverse function of log(x) is gol(x).

i.e. you can use the facts that d/dx log(x) = 1/x, log(1) = 0 (superfluous) and that log[gol(x)] = gol[log(x)] = x

EDIT: also that d/dx x = 1 etc.
y = gol(x)
log y = log [gol (x)] = x

Implicit differentiation

1/y dy/dx = 1
dy / dx = y = gol(x)

Why am I doing this anyway - it's summer
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Speleo
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#29
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Once you've done that:

Suppose that you have a function y(t), such that:
dy/dt = y.

Show that y(t) = gol(t). (use a definite integral preferably)
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Zii
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#30
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#30
(Original post by Speleo)
Using nothing more complicated than the chain rule, prove that the derivative of gol(x), the inverse function of log(x) is gol(x).

i.e. you can use the facts that d/dx log(x) = 1/x, log(1) = 0 (superfluous) and that log[gol(x)] = gol[log(x)] = x

EDIT: also that d/dx x = 1 etc.
y = x

\frac{\mathrm{d}y}{\mathrm{d}x} = 1

...

y = \mathrm{ln}(e^x) ( = x )

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{e^x} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(e  ^x)

...

\frac{1}{e^x} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(e  ^x) = 1

e^x = \frac{\mathrm{d}}{\mathrm{d}x}(e  ^x)

Q.E.D.
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Speleo
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#31
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Yup, it would be advisable to think about it as gol(x) or exp(x) [i.e. as the inverse function of log(x)] though, rather than e^x.
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Zii
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#32
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#32
(Original post by Speleo)
Once you've done that:

Suppose that you have a function y(t), such that:
dy/dt = y.

Show that y(t) = gol(t). (use a definite integral preferably)
If y = f(t) and \frac{\mathrm{d}y}{\mathrm{d}t} = y we get \frac{\mathrm{d}t}{\mathrm{d}y} = \frac{1}{y}.

\displaystyle\int^A_1 1 \, \mathrm{d}t = \displaystyle\int^A_1 \frac{1}{y} \, \mathrm{d}y

A-1 = \mathrm{ln}A

...and then it doesn't work?

I'm really not sure what to do.
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Cities
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#33
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#33
(Original post by Speleo)
Yup, it would be advisable to think about it as gol(x) or exp(x) [i.e. as the inverse function of log(x)] though, rather than e^x.
I solved it using a first order differential equation [using first principles]. What's your method?
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Speleo
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#34
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Hmm, the question is wrong but I think I can fix it up, give me a sec...
[exp(x-1) works as well for example]
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Speleo
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#35
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OK, the basic issue here is that when using separation of variables, we are implicitly integrating by substitution.

f(x) dx/dt = g(t)
INT f(x) dx/dt dt = INT g(t) dt
INT f(x) dx = INT g(t) dt by integration by substitution.

Now this means that we have to change the limits on the integral so that when x is at the upper limit in the x integral, t is at the upper limit in the t integral and similarly for the lower limits. i.e. if x = h(t), then x_max = h(t_max) and x_min = h(t_min).

Since we don't know how x relies on t, we have to use an indefinite integral instead.

Try it again, just the answer you should get should be gol(t+c) for a real number c.
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Zii
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#36
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(Original post by Speleo)
OK, the basic issue here is that when using separation of variables, we are implicitly integrating by substitution.

f(x) dx/dt = g(t)
INT f(x) dx/dt dt = INT g(t) dt
INT f(x) dx = INT g(t) dt by integration by substitution.

Now this means that we have to change the limits on the integral so that when x is at the upper limit in the x integral, t is at the upper limit in the t integral and similarly for the lower limits. i.e. if x = h(t), then x_max = h(t_max) and x_min = h(t_min).

Since we don't know how x relies on t, we have to use an indefinite integral instead.

Try it again, just the answer you should get should be gol(t+c) for a real number c.
Yeah I had to use an indefinite integral (is this because if you do a definite integral you can't have the same limits for different variables?)

I ended up with t + C = \mathrm{ln}y

y = \mathrm{exp}(t + C)

But how do I show that really, C ought to really be ?
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Speleo
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#37
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#37
You don't.

EDIT: I'm not really going anywhere with this...
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Zii
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#38
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#38
(Original post by Speleo)
You don't.
Is it because e^C is just a constant so whatever C is, it still satisfies f(x) = f '(x)?
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Speleo
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It's because d/dx (x+C) = 1. You don't know any properties of exp(x) at this point other than that it is the inverse of log(x) and is its own derivative. i.e. you don't know that e^(a+b) = e^a.e^b
You could try deriving this from log(ab) = log(a) + log(b) though.
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Cities
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#40
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(Original post by Speleo)
It's because d/dx (x+C) = 1. You don't know any properties of exp(x) at this point other than that it is the inverse of log(x) and is its own derivative. i.e. you don't know that e^(a+b) = e^a.e^b
You could try deriving this from log(ab) = log(a) + log(b) though.
Yeh I got the same with the first order diff eq. Anyway have fun guys. I may be drawn back here sometime in the near future lol
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