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# differentiation (hyperbolic ?) watch

1. Could anyone help me with question 4, I've been on it for an hour and got no clue how to do it.
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2. (Original post by ibrahim541)
Could anyone help me with question 4, I've been on it for an hour and got no clue how to do it.

What similarities can you see between and
3. (Original post by Khallil)

What similarities can you see between and
Y(x) = 2sinh(x) ?
4. (Original post by ibrahim541)
Y(x) = 2sinh(x) ?
Exactly

Do you know how to differentiate hyperbolic functions?
5. (Original post by Khallil)
Exactly

Do you know how to differentiate hyperbolic functions?
I do and after I differentiate it I get dy/dx=2cosh(x) but it wants it in terms of y and I've got no idea how to rearrange the original formula to get that
6. There's a useful identity:
7. (Original post by ibrahim541)
I do and after I differentiate it I get dy/dx=2cosh(x) but it wants it in terms of y and I've got no idea how to rearrange the original formula to get that
Ah I see. Have you learnt Osbourne's rule yet?

It states that to convert any trigonometric identity to a hyperbolic identity, you must change all cos terms into cosh and replace any product of two sin terms with -sinh^2 :

For example:

is the product of two terms so it is replaced by
8. Ahh okay, that helps a lot. so I got dx/dy=1/2(y^2+1)^0.5 can anyone clarify if this is right ?
9. (Original post by ibrahim541)
...

Use the fact that to simplify the expression for
10. (Original post by Khallil)

Use the fact that to simplify the expression for
Ah so its dx/dy=1/2(1+(y/2)^0.5)^0.5, thanks you so much for your help
11. (Original post by ibrahim541)
Ah so its dx/dy=1/2(1+(y/2)^0.5)^0.5, thanks you so much for your help
I think you mean:

12. (Original post by Khallil)
I think you mean:

Okay got it, thanks again

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