text: # Thread: Find The Derivative of the Function x^2√(1-x^2)
  
  1. ## Find The Derivative of the Function x^2√(1-x^2)
  
  f(x)=(x^2)(√(1-x^2))
  
  f'(x)=(2x)(√(1-x^2))+(x^2)(1/2)(√(1-x^2))^-1)(-2x)
  
  f'(x)=(2x)((1-x^2)^1/2)+(x^2)(1/2)(√(1-x^2)^-1)(-2x)
  
  f'(x)=√(4x^2-4x^4)+(1/2)(-2x^3)(1/(√(1-x^2))
  
  umm i'm lost, i probably didn't even get the above right (sorry the format sucks pretty bad)
  
  never mind i found a picture
  
  2. ## Re: Find The Derivative of the Function x^2√(1-x^2)
  
  $\displaystyle x^2\sqrt{1-x^2}$
  
  I'd use the product rule.
  
  Let $\displaystyle u=x^2$
  
  then $\displaystyle \frac{du}{dx}=?$
  
  Let $\displaystyle v=(1-x^2)^{\frac{1}{2}}$
  
  then $\displaystyle \frac{dv}{dx}=?$
  
  You should be able to proceed from here.
  
  3. ## Re: Find The Derivative of the Function x^2√(1-x^2)
  
  $\displaystyle x^2\sqrt{1-x^2}=\sqrt{x^4-x^6}$
  
  and
  
  $\displaystyle \frac{d}{dx}(\sqrt{f(x)})=\frac{f'(x)}{2\sqrt{f(x) }}$
  
  4. ## Re: Find The Derivative of the Function x^2√(1-x^2)
  
  Originally Posted by Deo3560
  f(x)=(x^2)(√(1-x^2))
  
  f'(x)=(2x)(√(1-x^2))+(x^2)(1/2)(√(1-x^2))^-1)(-2x)
  
  f'(x)=(2x)((1-x^2)^1/2)+(x^2)(1/2)(√(1-x^2)^-1)(-2x)
  $\displaystyle f'(x)=(2x)(1-x^2)^{1/2}-(x^2)(1/2)(1-x^2)^{-1/2}(-2x)$
  
  It is right up to here, and may be simplified:
  
  $\displaystyle f'(x)=(2x)(1-x^2)^{1/2}+\frac{x^3}{(1-x^2)^{1/2}}$
  
  It is easier to proceed by taking out a factor of $\displaystyle x \sqrt{1-x^2)$
  
  $\displaystyle f'(x) =x(1-x^2)^{1/2}\left[2 -\frac{x^2}{1-x^2} \right]$
  
  That square bracket may be simplified further if you wish.
  
  CB
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