f'(x) X/(x^2-4)^(1/2)

hi!! im stuck on this and dont understand how to find its derivitive. ive tried multipling by the conjugate and i cant get anything to cancel. it is an ap test question that my teacher gave us for homework.

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- Oct 15th 2009, 10:50 PMbromezf'(x)
f'(x) X/(x^2-4)^(1/2)

hi!! im stuck on this and dont understand how to find its derivitive. ive tried multipling by the conjugate and i cant get anything to cancel. it is an ap test question that my teacher gave us for homework. - Oct 15th 2009, 11:19 PMAmer
I will not solve it I will solve a similar example

$\displaystyle f(x) = \frac{\sin x }{(x^3-1)^{\frac{3}{2}}}$

I prefer to write it like this

$\displaystyle f(x) = (\sin x)(x^3-1)^{\frac{-3}{2}}$

$\displaystyle f'(x)= (x^3-1)^{\frac{-3}{2}}\left(\frac{d \sin x}{dx}\right) + (\sin x)\left(\frac{d(x^3-1)^{\frac{-3}{2}}}{dx}\right)$

$\displaystyle f'(x) = (x^3-1)^{\frac{3}{2}}(\cos x) + (\sin x)\left(\frac{-3}{2}(3x^2)(x^3-1)^{\frac{-5}{2}}\right)$

in general

$\displaystyle f(x) = g(x) h(x) $

$\displaystyle f'(x) = g'(x) h(x) + h'(x) g(x) $ - Oct 15th 2009, 11:38 PMVitaX
$\displaystyle f'(x)=x(x^2 - 4)^{-\frac{1}{2}}$

Use product and chain rule $\displaystyle y'=vu' + uv'$ where $\displaystyle v=(x^2 - 4)^{-\frac{1}{2}}$ and $\displaystyle u=x$

$\displaystyle u'=1$ and $\displaystyle v'=-\frac{1}{2}(x^2 - 4)^{-\frac{3}{2}}(2x)$

Substitute these values into the formula above.

$\displaystyle f'(x)=(x^2 -4)^{-\frac{1}{2}}(1) + (x)(-x(x^2 -4)^{-\frac{3}{2}})$ - Oct 15th 2009, 11:45 PMbromez
thx this should get me started

- Oct 15th 2009, 11:49 PMVitaX
Had an error with $\displaystyle v'$ above. Fixed it though