# What is the derivative of N(t) = 20 - 30/sqrt (9+t^2)?

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You have the function N(t) = 20 - 30/[sqrt(9 + t^2)]

N'(t) = [20]' - [30/[sqrt(9 + t^2)]]'

The derivative of a constant is 0, so the first term is 0, for the second term [30/[sqrt(9 + t^2)]]'

=> [30*(9 + t^2)^(-1/2)]'

=> 30*(-1/2)(9 + t^2)^(-3/2)*(2t)

=> -30t*(9 + t^2)^(-3/2)

**The derivative of N(t) = -30t*(9 + t^2)^(-3/2)**

Pretty sure I figured it out... Ignore the 20, it becomes 0 if you try to differentiate it.