## Calculus (3rd Edition)

$$y'= -2kt^3(kt^4+b)^{-3/2}.$$
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 $$y'=-\frac{1}{2}(kt^4+b)^{-3/2}(4kt^3)=-2kt^3(kt^4+b)^{-3/2}.$$