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

Limits, differentiation, related rates, integration, trig integrals, etc.

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

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.
arman

Posts: 1
Joined: Sun May 31, 2009 12:27 am

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.

stapel_eliz

Posts: 1756
Joined: Mon Dec 08, 2008 4:22 pm