i was looking at the question: x/(sqrt1 - x^2).
Normally i would think you need to use quotient property of differentials first then find the derivatives of the bottom term using the chain rule.
Could you split the original fraction into x/1 by 1/(sqrt1-x^2) and use the product rule for derivatives?
I attached a photo of what i would assume is the exact same function, would the derivatives be the same? im still having trouble with depravities so im not confident in myself solving both and being sure of the answers.
As I understand it, the question here is about how to find the
derivative of [imath]\frac{x}{\sqrt{1-x^2}}[/imath]. (Some of the answers look as if they were trying to find the
integral, though I could be wrong.)
Since a week has passed, let's go ahead and do this in the two suggested ways, in case anyone else needs to see it.
By the
quotient rule, [imath]\displaystyle\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{\frac{du}{dx}v-u\frac{dv}{dx}}{v^2}[/imath], where [imath]u=x[/imath] and [imath]v=\sqrt{1-x^2}=(1-x^2)^{1/2}[/imath].
Then [imath]\frac{du}{dx}=1[/imath] and [imath]\displaystyle\frac{dv}{dx}=\frac{1}{2}(1-x^2)^{-1/2}(-2x)=\frac{-x}{\sqrt{1-x^2}}[/imath].
So the derivative is [math]\frac{1\cdot\sqrt{1-x^2}-x\cdot\frac{-x}{\sqrt{1-x^2}}}{\left(\sqrt{1-x^2}\right)^2}=\frac{\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}}{1-x^2}\\=\frac{\sqrt{1-x^2}\cdot\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}\cdot\sqrt{1-x^2}}{\left(1-x^2\right)\cdot\sqrt{1-x^2}}=\frac{(1-x^2)+x^2}{\left(1-x^2\right)^{3/2}}\\=\frac{1}{\left(1-x^2\right)^{3/2}}[/math]
If we rewrite the function as [imath]x\cdot\frac{1}{\sqrt{1-x^2}}=x\cdot(1-x^2)^{-1/2}[/imath], then by the
product rule, [imath]\displaystyle\frac{d}{dx}\left(uv\right)=\frac{du}{dx}v+u\frac{dv}{dx}[/imath], where [imath]u=x[/imath] and [imath]v=(1-x^2)^{-1/2}[/imath], we get [imath]\frac{du}{dx}=1[/imath] and [imath]\frac{dv}{dx}=-\frac{1}{2}(1-x^2)^{-3/2}(-2x)=x(1-x^2)^{-3/2}[/imath].
Then the derivative is [math]1\cdot(1-x^2)^{-1/2}+x\cdot x(1-x^2)^{-3/2}=\frac{1}{\sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}^3}\\=\frac{1-x^2}{(1-x^2)\sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}^3}=\frac{1-x^2}{(1-x^2)^{3/2}}+\frac{x^2}{\left(1-x^2\right)^{3/2}}\\=\frac{1-x^2+x^2}{\left(1-x^2\right)^{3/2}}=\frac{1}{\left(1-x^2\right)^{3/2}}[/math]
The answers are the same; I think the second way is a little easier in some way. That's especially true if you don't need to simplify the answer.
