# If #(dx)/(dt) = "8 min"^(-1)# and #y = 1/x#, then what is #(dy)/(dt)# at #x = 2#?

##### 2 Answers

Let us consider the graph of

We are given

#x = x(t)# because the x value varies over time

#y = y(x(t))# because y varies as x varies, and x varies over time

So, we can write

#color(blue)((dy)/(dt)) = (dy(x(t)))/(dt)#

#= color(green)((dy(x(t)))/(dx(t))*(dx(t))/(dt))#

#= color(blue)((dy)/(dx)*(dx)/(dt))#

So, now let's actually take the derivative with respect to

#color(blue)((dy)/(dt)) = {:[-1/x^2(dx)/(dt)]|:}_(x=2)#

#= -1/(2)^2*("8 min"^(-1))#

#= color(blue)(-"2 min"^(-1))#

For