TITLE: First derivative of $\sqrt[\large 5]{\frac{t^3 + 1}{t + 1}}$
QUESTION [6 upvotes]: I have yet another derivative I need help with. I have to differentiate :
$$\sqrt[\uproot{3}{\Large 5}]{\frac{t^3 + 1}{t + 1}}$$
with respect to $t$. 
I had two thoughts about this, use the chain rule then the quotient rule and multiply out, but then I am left with a mess of:
$$\left(\frac{t^3+1}{t+1}\right)^{-4/5}\cdot\frac{0.6t^3+0.6t^2-0.2t^4+0.2t}{(t+1)^2}$$
This is turning into a real mess and the answer I should get is:
$$\frac{2t-1}{5(t^2-t+1)^{4/5}}$$
Am I going the correct way about this or should I try a different route?
Thanks

REPLY [4 votes]: I would prefer to take the fifth power of both sides and get
$$y^5(t+1)=t^3+1.$$
Differentiate both sides, using implicit differentiation on the left.
We get 
$$y^5+5(t+1)y^4 y'=3t^2.$$
Now solve for $y'$.
For more complicated examples of a similar kind, I would suggest logarithmic differentiation. Take the (natural) logarithm of both sides. We get
$$\log y=\frac{1}{5}\log(1+t^3)-\frac{1}{5}\log(1+t).$$
Differentiate both sides with respect to $t$. On the left we get $\frac{y'}{y}$. Differentiation on the right is straightforward.