# How do you differentiate g(t)=1/(t^4+1)^3?

To differentiate this function, you need to use the chain rule. When you do so, you get $g ' \left(t\right) = \frac{- 12 {t}^{3}}{{t}^{4} + 1} ^ 4$ .
Whenever you see a function nested inside another function—a composition of functions—and you want to find the derivative of the whole thing, you need to use the chain rule. In symbols, the chain rule states that $f \circ h = f ' \left(h\right) \cdot h '$. In this case, since the function ${t}^{4} + 1$ is then cubed and reciprocated, it is a composition of functions and the chain rule needs to be applied. $f \left(x\right) = {x}^{-} 3$ and $h \left(x\right) = {t}^{4} + 1$, so $g ' \left(x\right) = f ' \left({t}^{4} + 1\right) \cdot h ' = - 3 {\left({t}^{4} + 1\right)}^{-} 2 \cdot 4 {t}^{3} = \frac{- 12 {t}^{3}}{{t}^{4} + 1} ^ 4$.