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 $g^{\prime }(t)=-\frac{t}{t^2-2}$.

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 $g^{\prime }(t)=-\frac{f^{\prime }(t)}{[f(t){]}^2}$. So now we need to find $f^{\prime }(t)$.

Find $f^{\prime }(t)$

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 $f^{\prime }(t)$.

Apply the Chain Rule

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

Find $g^{\prime }(t)$

Substituting $f(t)$ and $f^{\prime }(t)$ back into the equation for $g^{\prime }(t)$ gives $g^{\prime }(t)=-\frac{f^{\prime }(t)}{[f(t){]}^2}=-\frac{t/\sqrt{t^2-2}}{(\sqrt{t^2-2}{)}^2}=-\frac{t}{t^2-2}$.