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?
 A: Let $$f(x)=\tan x$$ and $$g(x)=\frac{1}{1+x^2}$$ so $$f'(x)=1+\tan^2 x$$ and $$g'(x)=\frac{-2x}{(1+x^2)^2}$$ and you want differentiate $f(g(x))$ so use the chaine rule:
$$(f\circ g)'=(f'\circ g)\times g'$$
A: 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: $$\dfrac d{dx}\left(\tan \left(\frac{1}{ x^2 +1}\right)\right) = \dfrac{d}{dx}\Big(\tan(f(x)\Big)\cdot \Big(f'(x)\Big) = \sec^2 \left(\dfrac 1{1 + x^2}\right)\cdot \left(-\dfrac{2x}{(1+ x^2)^2}\right)$$
A: $\displaystyle \frac d {dx}\tan\frac 1 {1+x^2}$
$\displaystyle=\left(\frac d {du} \tan u\right)\left(\frac d {dx}\frac 1 {1+x^2}\right)$ [Chain rule]
$=\cdots$ [Quotient rule]
A: Answering this one too now, I'll write the steps straight away
$$\frac{d}{dx}\left(\tan \frac{1}{1+x^2} \right)=
\frac{d}{dy}\left.\left(\tan y \right)\right|_{y=\frac{1}{1+x^2}}
\cdot\frac{d}{dx}\left( \frac{1}{1+x^2} \right)$$
$$=\sec^2y|_{y=\frac{1}{1+x^2}} \cdot \left( - \frac{2x}{1+x^2} \right)$$
$$=\sec^2 \left( \frac{1}{1+x^2} \right) \cdot \left( - \frac{2x}{1+x^2} \right)$$
A: Your function is not a quotient. It is the tangent of a quotient, in other words $f(x) = \tan g(x)$ or equivalently $f = \tan \circ \, g$. It is a composition of functions. If you have been given this exercise then you must have been taught how to differentiate a composition: $(u\circ v)' = (u' \circ\, v) \cdot v'$.
