Question

find the differential of ( y=sqrt{8 + t^{2}} ).
( \frac{dy}{dt}=square )

Explanation:

Step1: Rewrite the function

Rewrite \(y = \sqrt{8 + t^{2}}\) as \(y=(8 + t^{2})^{\frac{1}{2}}\).

Step2: Apply the chain rule

The chain rule states that if \(y = f(g(t))\), then \(\frac{dy}{dt}=f^{\prime}(g(t))\cdot g^{\prime}(t)\). Let \(u = 8 + t^{2}\), so \(y = u^{\frac{1}{2}}\). First, find \(\frac{dy}{du}\): \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}\). Then find \(\frac{du}{dt}\): \(\frac{du}{dt}=2t\).

Step3: Calculate \(\frac{dy}{dt}\)

By the chain rule \(\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}\). Substitute \(u = 8 + t^{2}\), \(\frac{dy}{du}=\frac{1}{2}(8 + t^{2})^{-\frac{1}{2}}\) and \(\frac{du}{dt}=2t\) into the formula: \(\frac{dy}{dt}=\frac{1}{2}(8 + t^{2})^{-\frac{1}{2}}\cdot2t\). Simplify the expression: \(\frac{dy}{dt}=\frac{t}{\sqrt{8 + t^{2}}}\).

Answer:

\(\frac{t}{\sqrt{8 + t^{2}}}\)