derivative involving ln

[Math wiz ya rite 09]

For |x| < 1, the derivative of y = ln*sqrt(1-x^2) is...

A. x/(1 - x^2)
B. x/(x^2 -1)
C. -x/(x^2 -1)
D. 1/(2*(1-x^2))
E. 1/(sqrt(1 - x^2))

 

[arthur ohlsten]

y=ln [1-x^2]^1/2 l x l <1
y=1/2 ln [1-x^2]
dy/dx = [1/2] 1/[1-x^2][-2x]
dy/dx= x/[x^2-1] answer B

Arthur

 

[Math wiz ya rite 09]

shoulnd't the answer be see if you did proper steps. because you get

(-2x) / [2(1-x^2)]

the negative does nto reduce

only the 2's

can i get confirmation on this problem.

thanks

 

[jwpaine]

$\displaystyle f(x) = ln\sqrt(1-x^2)$
$\displaystyle f'(x) = \frac{d}{dx}[ln\sqrt(1-x^2)]$

(Use chain rule twice)

$\displaystyle f'(x) = [\sqrt{1-x^2} \rightarrow \frac{1}{x}][\frac{d}{dx}\sqrt{1-x^2}]$
$\displaystyle f'(x) = [\sqrt{1-x^2} \rightarrow \frac{1}{x}][\,\,[(1-x^2) \rightarrow \frac{1}{2\sqrt{x}}][\frac{d}{dx}(1-x^2)]$
$\displaystyle f'(x) = [\sqrt{1-x^2} \rightarrow \frac{1}{x}][\,\,[(1-x^2) \rightarrow \frac{1}{2\sqrt{x}}][-2x]$
$\displaystyle f'(x) = [\frac{1}{\sqrt{1-x^2}}][\frac{1}{2\sqrt{1-x^2}}][-2x]$
$\displaystyle f'(x) = \frac{-2x}{2(1-x^2)}$
$\displaystyle f'(x) = \frac{-x}{(1-x^2)}$
$\displaystyle f'(x) = \frac{x}{(x^2-1)}$

we can bring the -1 from the numerator to the denominator: $\displaystyle -(1-x^2) = (-1 + x^2) = (x^2 - 1)$



Cheers,
John