Find the derivative using limit definition: $\frac{x}{1 + x\sin(1/x)}$ 
How can I find the derivative of this function  (defined below) using the definition of the limit?
$$f(x):= \frac{x}{1 + x\sin(1/x)}$$

So I know the definition of the limit but I am unable to apply it here. Any help or hint would be appreciated.
 A: Assuming that you know that $$\frac{df}{dx}=\frac{df}{dy}\times \frac{dy}{dx}$$ the problem becomes simple making $x=\frac 1y$ that is to say $y=\frac 1x$ and $\frac{dy}{dx}=-\frac 1 {x^2}$.
So $$f(y)=\frac{1}{y+\sin (y)}$$ $$f(y+h)-f(y)=\frac{1}{(y+h)+\sin (y+h)}-\frac{1}{y+\sin (y)}$$
$$f(y+h)-f(y)=\frac{y+\sin (y)-(y+h)-\sin (y+h)}{\left(y+\sin (y) \right) \left((y+h)+\sin (y+h) \right)}$$ $$f(y+h)-f(y)=-\frac{h+\sin (y+h)-\sin (y)}{\left(y+\sin (y) \right) \left((y+h)+\sin (y+h) \right)}\tag 1$$ Now, use
$$\sin (y+h)-\sin (y)=2 \sin\left(\frac{y+h-y }2\right)\cos\left(\frac{y+h+y }2\right)=2\sin\left(\frac{h }2\right)\cos\left(y+\frac{h }2\right)$$ making
$$h+\sin (y+h)-\sin (y)=2\left(\frac h2+\sin\left(\frac{h }2\right)\cos\left(y+\frac{h }2\right)\right)$$ $$h+\sin (y+h)-\sin (y)=2\times\frac h2\left(1+\frac{\sin\left(\frac{h }2\right)}{\frac h 2}\cos\left(y+\frac{h }2\right)\right)\to h\left(1+\cos\left(y\right)\right)$$ Back to $(1)$ we then have $$f(y+h)-f(y)\to -\frac {h\left(1+\cos\left(y\right)\right)}{(y+\sin(y))^2}$$ which makes $$\frac{df}{dy}=-\frac {\left(1+\cos\left(y\right)\right)}{(y+\sin(y))^2}$$ Replace now $y$ by $\frac 1x$ and write, as said at the beginning $$\frac{df}{dx}=\frac{df}{dy}\times \frac{dy}{dx}=-\frac 1{x^2}\times \frac{df}{dy}$$ and simplify.
