Finding the derivative of a function. Please check my answer Find the derivative of
$$f(x)=x^2 \sqrt{1-x^2}$$
Here is my solution .
 A: Lets analyze what you did. The problem is to find the derivative of 
$$f(x)=x^2 \sqrt{1-x^2}.$$ From the 2nd line of your solution, you wrote
$$f'(x)=2x\bigg[\frac{1}{2}(1-x^2)^{-\frac{1}{2}} \bigg]\cdot D_x(1-x^2).$$ It seems that your interpretation of the derivative of the product $uv$, where $u$ and $v$ are both functions of $x$, is $u'v'$, that is, you are thinking that $$(uv)'=u'v'.$$ This is WRONG! It should be $$(uv)'\quad=\quad u\cdot v'\quad+\quad v\cdot u'.\tag{formula}$$ So, to get a correct solution, follow the following steps:
Step 1: Let $u=x^2$ and $v=\sqrt{1-x^2}$.
Step 2. Find $u'$ and $v'$.
Step 3. Then use step 2 to get
$$\begin{align}f'(x)&=(uv)'\\
&=u\cdot v'\quad+\quad v\cdot u'\\
&=
\end{align}$$
A: Just another way to do the work when you face products, quotients and powers :  logarithmic differentiation $$f=x^2 \sqrt{1-x^2}\implies \log(f)=2\log(x)+\frac12 \log(1-x^2)$$ Differentiate $$\frac {f'}f=\frac 2x-\frac x{1-x^2}=\frac{2-3x^2}{x(1-x^2)}$$ $$f'=f \times \frac {f'}f=x^2 \sqrt{1-x^2}\times\frac{2-3x^2}{x(1-x^2)}=\frac{x(2-3x^2)}{\sqrt {1-x^2}}$$
