find derivative when y=x^3/1-x^2

Question

myininaya · Answer

$$y=\frac{f}{g} , g \neq 0 => y'=\frac{f' g- f g'}{(g)^2}$$

anonymous · Answer

Use the product (or quotient if you care to remember it) and chain rules$$y=x^{3}(1-x^{2})^{-1}$$$${dy\over dx}=x^{3}{d\over dx}(1-x^{2})^{-1}+(1-x^{2})^{-1}{d\over dx}x^{3}$$$${dy\over dx}={x^{3}(2x)\over (1-x^{2})^{2}}+{3x^{2}\over (1-x^{2})^{1}}$$Simplify if you need to (I won't because it is a pain to keep writing all the equation code)

Also, as a note, the chain rule was applied to take the derivative of (1-x^2)^-1$${d\over dx}(1-x^{2})^{-1}, \text{let u}=1-x^{2}, {du\over dx}=-2x$$$${d\over dx}(u)^{-1}=-u^{-2}{du\over dx}={2x\over (1-x^{2})^{2}}$$