Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ s(t)=\frac{1}{t^{2}+3 t-1} $$

Short answer

The derivative of the given function \(s(t)=\frac{1}{t^{2}+3t-1}\) is \( ds/dt = -1*(2t + 3) /((t^{2}+3t-1)^{2})\). The Chain Rule and the Power Rule of differentiation were used to derive this solution.

Step-by-step solution

Rewrite Function

Rewrite function in proper format i.e., \(s(t)= (t^{2}+3t-1)^{-1}\). This format is more suitable for differentiation because it allows us to use the power rule.

Apply Chain Rule

Differentiate using Chain Rule which is a method used for differentiating a function of a function. We can do this by differentiating the outside function and then multiplying that by the derivative of the inside function. Chain rule of differentiation: \(dy/dx = dy/du * du/dx\). Here, let \( u = t^{2}+3t-1\) , so \( du/dt = 2t + 3\). Applying chain rule we get, \( ds/dt = -1*(t^{2}+3t-1)^{-2} * (2t + 3) \).

Simplify Answer

Simplify the answer to get the final derivative. Multiply the exponents together to simplify the expression and we get \( ds/dt = -1*(2t + 3) /((t^{2}+3t-1)^{2})\). This is the derivative of the given function.