# f'(x)

• Oct 15th 2009, 11:50 PM
bromez
f'(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 16th 2009, 12:19 AM
Amer
Quote:

Originally Posted by bromez
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.

I will not solve it I will solve a similar example

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

I prefer to write it like this

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

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

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

in general

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

$f'(x) = g'(x) h(x) + h'(x) g(x)$
• Oct 16th 2009, 12:38 AM
VitaX
Quote:

Originally Posted by bromez
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.

$f'(x)=x(x^2 - 4)^{-\frac{1}{2}}$
Use product and chain rule $y'=vu' + uv'$ where $v=(x^2 - 4)^{-\frac{1}{2}}$ and $u=x$
$u'=1$ and $v'=-\frac{1}{2}(x^2 - 4)^{-\frac{3}{2}}(2x)$
Substitute these values into the formula above.
$f'(x)=(x^2 -4)^{-\frac{1}{2}}(1) + (x)(-x(x^2 -4)^{-\frac{3}{2}})$
• Oct 16th 2009, 12:45 AM
bromez
thx this should get me started
• Oct 16th 2009, 12:49 AM
VitaX
Had an error with $v'$ above. Fixed it though