- #1

- 171

- 0

does anyone know how to get this/derive this i.e. not take from tables !!!!

Thanks

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- Thread starter coverband
- Start date

- #1

- 171

- 0

does anyone know how to get this/derive this i.e. not take from tables !!!!

Thanks

- #2

- 499

- 2

y = Sqrt[x^2 - a]

y = [x^2 - a]^(1/2)

Take natural log

Ln[y] = Ln{ [x^2 - a]^(1/2) }

Ln[y] = (1/2) Ln{[x^2 - a]}

Take derivative wrt x. Don't forget you must apply chain rule to right hand side. Where y' comes out anyway.

(1/y) (y') = 2x/ (2 (x^2 - a))

writing more neatly, cancel a two

(y'/y) = x / (x^2 - a)

Finally, multiply a y back over and resub your origional y.

y' = yx/ (x^2 - a) y = [x^2 - a]^(1/2)

Thus

y' = x [x^2 - a]^(1/2) / (x^2 - a)

y' = x(x^2 - a)^(-1/2)

- #3

- 499

- 2

I'm an idiot, I took a deravitive. Not even the easiest way either.

- #4

Defennder

Homework Helper

- 2,591

- 5

Use a trigo substitution. Specifically, look at the ones for sec and tan.

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