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To start with I've worked out the drivatives of the two functions, getting 1 over x sqrt(x^(2) - 1) and the same thing again only the negaive respectively. Can you give me a hint as to where I should be going with this question? Thanks!

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To start with I've worked out the drivatives of the two functions, getting 1 over x sqrt(x^(2) - 1) and the same thing again only the negaive respectively. Can you give me a hint as to where I should be going with this question? Thanks!

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matt grime

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then you know df/dx=-dg/dx

that is, using that only constant functions have zero derivative,

f=-g+k for some constant k, put in some nice value of x to find what k is

sec(pi/4)= sqrt(2) so pi/4 = arcseec(sqrt(2))

and cot(pi/4)=1 so arccot(1)=pi/4

hence setting x=sqrt2 in the equation f=-g+k says that

pi/4=-pi/4+k, ie k=pi/2

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HallsofIvy

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In other words, if (f-g)'= f'- g'= 0 for all x (or all x on an interval) then f-g is a constant (on that interval). If you can show further that, for

When you say "To start with I've worked out the drivatives of the two functions, getting 1 over x sqrt(x^(2) - 1) and the same thing again only the negative respectively", what two functions are you talking about? I think you mean arcsec(x) and arccot(sqrt(x^(2) -1) but the two functions you want to prove equal are arcsec(x) and

pi/2-arccot(sqrt(x^(2) -1). Of course, the derivative of

pi/2-arccot(sqrt(x^(2) -1), because of that "- arccot.." IS precisely the negative of the derivative of arccot(sqrt(x^(2) -1) so you have in fact, shown that the two functions have the same derivative and, so, differ by some constant.

Now if you can show that those two functions have the same value at, say, sqrt(2) (because that's particularly easy!) you will have shown that that constant is 0 and the two functions are equal for all x.

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matt grime

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Originally posted by Claire84

df/dx=-dg/dx

so rearranging

d?dx(f+g)=0

implies f+g=k some constant k so

f=-g+k

in HoI method it is, I think, that h=pi/2-arccot(junk)

and he shows that df/dx=dh/dx, and that h(sqrt(2))=f(sqrt(2))

I chose mine because you said

"To start with I've worked out the drivatives of the two functions, getting 1 over x sqrt(x^(2) - 1) and the same thing again only the negaive respectively"

so I labelled the functions accordingly, HoI took a different reading of that, that's just where the difference comes from. Nothing special, just that it isn't clear what 'ther two functions' were that you were talking about.

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HallsofIvy

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That "HoI" stopped me for a moment. I have a friend who, on the internet, uses the charming sobriquet "HogOnIce" and he regularly abbreviates it HOI! It hadn't occured to me that "HallsofIvy" works out the same way.

By the way, matt, I took the "two functions" Clair84 had differentiated to be only arcsec(x) and 1 over x sqrt(x^(2) - 1) and the same thing again only the negaive respectively (rather than pi/2-arccot(sqrt(x^(2) -1)) precisely because, that way, the derivatives are "1 over x sqrt(x^(2) - 1) and the same thing again only the negative respectively". Because of that, arcsec(x) and pi/2-arccot(sqrt(x^(2) -1)) have the same derivatives.

Clair84- "Right, can you just explain the bit about the zero derivative please, like how you can get to f = -g +k ?"

I didn't say anything about that because that is exactly what you "hint" said. It can be proven that "If f'(x)= 0 at every point on an interval then f(x) is a constant" using the mean value theorem.

Recall that the mean value theorem says "If f is continuous on the interval [a,b] and differentiable on (a,b) (in other words f has to be continuous at the two endpoints but not necessarily differentiable there) then there exist c in (a,b) such that

(f(b)- f(a))/(b-a)= f'(c)."

Assume f'(x)= 0 at every point in an interval (so, in particular, it is differentiable at every point). Take a, b to be any two points in that interval and apply the mean value theorm: (f(b)- f(a))/(b-a)= f'(c)= 0 so f(b)- f(a)=0 and f(b)= f(a). Since a, b could be any two points, f is a constant function on the interval.

Now suppose f and g are two functions such that f'(x)= g'(x) at every point of an interval. Then f-g is a differentiable function and (f-g)'= f'-g'= 0. That is, f-g is a constant: f-g= C so f= g+ C.

By the way, matt, I took the "two functions" Clair84 had differentiated to be only arcsec(x) and 1 over x sqrt(x^(2) - 1) and the same thing again only the negaive respectively (rather than pi/2-arccot(sqrt(x^(2) -1)) precisely because, that way, the derivatives are "1 over x sqrt(x^(2) - 1) and the same thing again only the negative respectively". Because of that, arcsec(x) and pi/2-arccot(sqrt(x^(2) -1)) have the same derivatives.

Clair84- "Right, can you just explain the bit about the zero derivative please, like how you can get to f = -g +k ?"

I didn't say anything about that because that is exactly what you "hint" said. It can be proven that "If f'(x)= 0 at every point on an interval then f(x) is a constant" using the mean value theorem.

Recall that the mean value theorem says "If f is continuous on the interval [a,b] and differentiable on (a,b) (in other words f has to be continuous at the two endpoints but not necessarily differentiable there) then there exist c in (a,b) such that

(f(b)- f(a))/(b-a)= f'(c)."

Assume f'(x)= 0 at every point in an interval (so, in particular, it is differentiable at every point). Take a, b to be any two points in that interval and apply the mean value theorm: (f(b)- f(a))/(b-a)= f'(c)= 0 so f(b)- f(a)=0 and f(b)= f(a). Since a, b could be any two points, f is a constant function on the interval.

Now suppose f and g are two functions such that f'(x)= g'(x) at every point of an interval. Then f-g is a differentiable function and (f-g)'= f'-g'= 0. That is, f-g is a constant: f-g= C so f= g+ C.

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