Answer
$$ y'= -2kt^3(kt^4+b)^{-3/2}.$$
Work Step by Step
Recall that $(x^n)'=nx^{n-1}$
Since $ y=\frac{1}{ \sqrt{kt^4+b}}$, we rewrite $ y $ as follows
$$ y=(kt^4+b)^{-1/2}.$$
Now, by using the chain rule, the derivative $ y'$ is given by
$$ y'=-\frac{1}{2}(kt^4+b)^{-3/2}(4kt^3)=-2kt^3(kt^4+b)^{-3/2}.$$
