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# How do you prove that the derivative of tanh^-1(x/a) is a/(a^2-x^2)? Watch

1. Yeah...question above says it all

Can't find how to do it in any books
2. do you mean from first principles?

if not, let y=artanh(x/a)

then tanh(y)=x/a

differentiate and rearrange for dy/dx
4. (Original post by Pheylan)
do you mean from first principles?

if not, let y=artanh(x/a)

then tanh(y)=x/a

differentiate and rearrange for dy/dx
Could you explain that a bit further...I know that differentiating tanh(y) would give sech^2(y) but I get stuck after that
5. How does sech^2 relate to tanh?
6. I am an idiot.
7. (Original post by IrrationalNumber)
How does sech^2 relate to tanh?
It's the derivative of it
8. (Original post by kellywellydoodle)
It's the derivative of it
True, let me try again
How does sech^2 relate to tanh^2?
9. (Original post by StephenP91)
Well.

Differentiate that.

tanh not tan
10. (Original post by StephenP91)
Well.

Differentiate that.

These are hyperbolic functions Stephen.
11. (Original post by Mr M)
These are hyperbolic functions Stephen.
Just noticed. You think'd I'd realise that there is a h infront of the functions. Derp.
12. what id do from my little knowledge of maths is draw the graph and work out the gradient at points, but im a bit of a noob, forgive me
13. (Original post by StephenP91)
Just noticed. You think'd I'd realise that there is a h infront of the functions. Derp.
As it happens, your working provides a pretty good hint on how to proceed.
14. (Original post by kellywellydoodle)
Yeah...question above says it all

Can't find how to do it in any books
Couple of ways. You can integrate a/(a^2 - x^2) and get tanh^-1(x/a) + k (best use a hyperbolic substitution). Hence in the case k = 0 this implies d/dx [tanh^-1(x/a)] = a/(a^2 - x^2). Or, differentiate directly as follows:

Let y = tanh^-1(x/a) ... (thus require dy/dx)
=> tanhy = x/a
=> sech^2y(dy/dx) = 1/a
=> dy/dx = 1/[asech^2y]

We need to replace sech^2y with a function of x and get the required result.
tanhy = x/a => 1 - tanh^2y = 1 - x^2/a^2 => sech^2y = 1 - x^2/a^2 (relate tanh^2y to sech^2y using the identity)

Replace sech^2y with the expression above. Fiddle a bit and voila.
15. (Original post by Mr M)
As it happens, your working provides a pretty good hint on how to proceed.
Borderline full solution to be honest.
16. (Original post by Pheylan)
do you mean from first principles?

if not, let y=artanh(x/a)

then tanh(y)=x/a

differentiate and rearrange for dy/dx
I think the question requires it from first principles actually
17. (Original post by kellywellydoodle)
I think the question requires it from first principles actually
use the definition of the derivative
18. (Original post by kellywellydoodle)
I think the question requires it from first principles actually
If it required it from first principles then it would definitely say so. Otherwise, you can use any method you like. Check out my earlier post.

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