Question

Finding a Derivative In Exercises \(7-34,\) find the derivative of the function. $$ g(t)=\frac{1}{\sqrt{t^{2}-2}} $$

Short answer

The derivative of the function \( g(t) = \frac{1}{\sqrt{t^{2}-2}} \) is \( g'(t) = -\frac{t}{t^{2}-2} \).

Step-by-step solution

Identify the Outer Function

The outer function can be seen as \( g(t) = \frac{1}{f(t)} \) where \( f(t) = \sqrt{t^{2}-2} \). So, to begin with, the power rule for differentiation has to be applied.

Apply the Power Rule

The power rule states that if \( g(t) = \frac{1}{f(t)} \) then its derivative \( g'(t) = -\frac{f'(t)}{[f(t)]^{2}} \). So now we need to find \( f'(t) \).

Find \( f'(t) \)

The function \( f(t) = \sqrt{t^{2}-2} \) can be rewritten as \( f(t) = (t^{2}-2)^{\frac{1}{2}} \). Now we can apply the chain rule to find \( f'(t) \).

Apply the Chain Rule

The chain rule states: If \( f(t) = (u(t))^n \), then \( f'(t) = n*(u(t))^{n-1}*u'(t) \). Here, \( u(t) = t^{2}-2 \), and it's derivative \( u'(t) = 2t \). Applying the chain rule gives \( f'(t) = \frac{1}{2}*(t^{2}-2)^{-\frac{1}{2}}*2t = \frac{t}{\sqrt{t^{2}-2}} \).

Find \( g'(t) \)

Substituting \( f(t) \) and \( f'(t) \) back into the equation for \( g'(t) \) gives \( g'(t) = -\frac{f'(t)}{[f(t)]^{2}} = -\frac{t/\sqrt{t^{2}-2}}{(\sqrt{t^{2}-2})^{2}} = -\frac{t}{t^{2}-2} \).