find the differential of ( y=sqrt{8 + t^{2}} ).\n( \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}}})