Differentiating $\tan\left(\frac{1}{ x^2 +1}\right)$ Differentiate: $\displaystyle \tan \left(\frac{1}{x^2 +1}\right)$
Do I use the quotient rule for this question? If so how do I start it of?

We use the chain rule to evaluate  $$ \dfrac{d}{dx}\left(\tan \frac{1}{x^2 +1}\right)$$
Since we have a function which is a composition of functions: $\tan(f(x))$, where $f(x) = \dfrac 1{1+x^2}$, this screams out chain-rule!
Now, recall that $$\dfrac{d}{dx}(\tan x) = \sec^2 x$$ 
and to evaluate $f'(x) = \dfrac d{dx}\left(\dfrac 1{1 + x^2}\right)$, we can use either the quotient rule, or the chain rule. Using the latter, we have $$\dfrac{d}{dx}\left(\dfrac 1{1 + x^2}\right)= \dfrac{d}{dx}(1 + x^2)^{-1} = -(1 + x^2)^{-2}\cdot \dfrac d{dx}(1+ x^2) = -\dfrac{2x}{(1+ x^2)^2}$$
Now, we put the "chain" together: $$\dfrac d{dx}\left(\tan \left(\frac{1}{ x^2 +1}\right)\right) = \dfrac{d}{dx}\Big(\tan(f(x)\Big)\cdot \Big(f'(x)\Big) = \sec^2 \left(\dfrac 1{1 + x^2}\right)\cdot \left(-\dfrac{2x}{(1+ x^2)^2}\right)$$
