First derivative of $\sqrt[\large 5]{\frac{t^3 + 1}{t + 1}}$ 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
 A: The process you describe one way to approach this: you can  use both the chain rule and the quotient rule. But it looks to me as though you misunderstand the quotient rule:
$$h(x) = \frac{f(x)}{g(x)}$$
$$h'(x) = \frac{f'(x)\cdot g(x) - f(x)\cdot g'(x)}{g^2(x)}$$
where the factors in the numerator are multiplied, not added. (It also looks like you multiplied the coefficients by $1/5$?)
In your case, the rational function within the 5th root is $$h(x) = \frac{f(x)}{g(x)} = \frac{t^3 + 1}{t+1}$$
$$f(x) = t^3 + 1, \implies f'(x) = 3t^2;\quad g(x) = t+ 1\implies g'(x) = 1$$
$$h'(x) = \frac{f'(x)\cdot g(x) - f(x)\cdot g'(x)}{g^2(x)} $$
$$h'(x) = \frac{3t^2(t+1) - (t^3 +1)\cdot 1}{(t+1)^2}\;\;=\;\;\frac{2t^3 + 3t^2 - 1}{(t+1)^2}$$
Now, multiply $h'(x)$ with the result you got for the first factor: $$\frac 15\left(\frac{t^3+1}{t+1}\right)^{-4/5}= \left(\frac{t+1}{t^3 + 1}\right)^{4/5}$$
Which gives us:
$$\frac 15\left(\frac{t+1}{t^3 + 1}\right)^{4/5}\cdot \frac{2t^3 + 3t^2 - 1}{(t+1)^2}$$
Now, we can simplify.  (The Chaz's method greatly simplifies the process of taking the derivative of your function. But I think it is a good idea to ensure you understand the quotient rule.)
Chaz's tip will also simplify the left factor above to $\left(\dfrac{1}{t^2 - t + 1}\right)^{4/5}$. After that, there is really no need to simplify. Answers can be correct, yet not match the "textbook's solution" perfectly, simply based on how we choose to simplify.
$$\left(\frac{2t - 1}{5(t^2 + t - 1)^{4/5}}\right)$$
A: Hint: simplify the radicand to $t^2 -t + 1$ before differentiating. 
A: 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. 
