The Purplemath Forums

by **arman** on Sun May 31, 2009 10:24 pm

y= x/sqrt[x²-4]

y'= (1)(sqrt[x²-4])-(x)[(1/2)(x²-4)^-1/2(2x)]/(x²-4)
= (sqrt[x²-4])-{(x)[(x)(x²-4)^-1/2]}/(x²-4)
= (sqrt[x²-4])-[(x²)(x³-4x)^-1/2]/(x²-4)
= [(sqrt[x²-4])-x²]/[(x²-4)(sqrt[x³-4x])]

I also have to do the second derivative, but I would just like to make sure the first is right before continuing. Thanks.

**Sponsor**

by **stapel_eliz** on Mon Jun 01, 2009 12:17 pm

arman wrote:y= x/sqrt[x²-4]

y'= (1)(sqrt[x²-4])-(x)[(1/2)(x²-4)^-1/2(2x)]/(x²-4)
= (sqrt[x²-4])-{(x)[(x)(x²-4)^-1/2]}/(x²-4)
= (sqrt[x²-4])-[(x²)(x³-4x)^-1/2]/(x²-4)

You can't do what you're trying to do with the radicals. The product of x and sqrt[x² - 4] is xsqrt[x² - 1] or sqrt[x⁴ - 4x²], since x = sqrt[x²] (for non-negative x); x does not equal sqrt[x].

You applied the Quotient Rule correctly. Then:

. . . . . $\frac{\sqrt{x^2\, -\, 4}\, -\, \frac{x^2}{\sqrt{x^2\, -\, 4}}}{\left(\sqrt{x^2\, -\, 4}\right)^2}$

. . . . . $\frac{\frac{x^2\, -\, 4}{\sqrt{x^2\, -\, 4}}\, -\, \frac{x^2}{\sqrt{x^2\, -\, 4}}}{\left(\sqrt{x^2\, -\, 4}\right)^2}$

. . . . . $\frac{\frac{-4}{\sqrt{x^2\, -\, 4}}}{\left(\sqrt{x^2\, -\, 4}\right)^2}$

. . . . . $\frac{-4}{\left(x^2\, -\, 4\right)^{\frac{3}{2}}$

Now differentiate again to get the second derivative.

The Purplemath Forums

Please check my 1st derivative? y = x/sqrt[x^2 - 4]

Please check my 1st derivative? y = x/sqrt[x^2 - 4]