On p. 34 of Schaum's Outline, Calculus, Frank Ayres, problem 40 gives the derivative of of f(x)=(x-1/x=1)^1/2 as I/(x=1)(x^2-1)^1/2.

I get as far as ((x-1/x=1)^-1/2)/(x=1)^2 but cannot reduce it to the answer given. Can someone help.

Thanks

MMH

On p. 34 of Schaum's Outline, Calculus, Frank Ayres, problem 40 gives the derivative of of f(x)=(x-1/x=1)^1/2 as I/(x=1)(x^2-1)^1/2.

I get as far as ((x-1/x=1)^-1/2)/(x=1)^2 but cannot reduce it to the answer given. Can someone help.

Thanks

MMH

I get as far as ((x-1/x=1)^-1/2)/(x=1)^2 but cannot reduce it to the answer given. Can someone help.

Thanks

MMH

- stapel_eliz
**Posts:**1670**Joined:**Mon Dec 08, 2008 4:22 pm-
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mmh28 wrote:...the derivative of of f(x)=(x-1/x=1)^1/2....

For what mathematical operation is your book using the second "equals" sign above?

Also, your

. . . . .

Was the above what you meant, or did you mean something more like the below?

. . . . .

Thank you!

Oops! I just discovered that the plus sign on my numeric pad prints out an equal sign. Weird.

I'll try again:http:

f(x) = ((x-1)/(x+1))^1/2

f'(x) = ((x-1)/(x+1))^-1/2)/(x+1)^2 (my answer, which I think is correct but can't get it in the form given in the book).

Answer in Schaum's: f'(x) = I/(x+1)(x^2-1)^1/2

Your second example, but with x+1 in the denominator, is what I intended.

I'm new to the forum but can't find out how to format with the actual radical sign as you have.

Many thanks

MMH

I'll try again:http:

f(x) = ((x-1)/(x+1))^1/2

f'(x) = ((x-1)/(x+1))^-1/2)/(x+1)^2 (my answer, which I think is correct but can't get it in the form given in the book).

Answer in Schaum's: f'(x) = I/(x+1)(x^2-1)^1/2

Your second example, but with x+1 in the denominator, is what I intended.

I'm new to the forum but can't find out how to format with the actual radical sign as you have.

Many thanks

MMH

- stapel_eliz
**Posts:**1670**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

mmh28 wrote:Answer in Schaum's: f'(x) = I/(x+1)(x^2-1)^1/2

What is meant by the "pipe" character in the numerator of the above? Is the radical in the numerator (multiplied against the fraction, and thus ending up on top) or in the denominator?

Note: I get the first step in the derivative as being:

. . . . .

This simplifies as:

. . . . .

Reduce, note that (x + 1)

Thank you! It was driving me mad.

MMH

MMH