Question

linearization and differentials: pr
(1 point)
find the differential of ( y=sqrt{10 + t^{2}} ).
( d y=square d t )
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Explanation:

Step1: Rewrite the function

Rewrite \(y = \sqrt{10 + t^{2}}=(10 + t^{2})^{\frac{1}{2}}\).

Step2: Apply the chain - rule

The chain - rule states that if \(y = f(g(t))\), then \(y^\prime=f^\prime(g(t))\cdot g^\prime(t)\). Let \(u = 10 + t^{2}\), so \(y = u^{\frac{1}{2}}\).
First, find \(\frac{dy}{du}\): \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}\).
Second, find \(\frac{du}{dt}\): \(\frac{du}{dt}=2t\).
Then \(\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}\). Substitute \(u = 10 + t^{2}\) into \(\frac{dy}{du}\), we get \(\frac{dy}{dt}=\frac{1}{2\sqrt{10 + t^{2}}}\cdot2t\).

Step3: Simplify the expression

\(\frac{dy}{dt}=\frac{t}{\sqrt{10 + t^{2}}}\).
Since \(dy=\frac{dy}{dt}dt\), so \(dy=\frac{t}{\sqrt{10 + t^{2}}}dt\).

Answer:

\(\frac{t}{\sqrt{10 + t^{2}}}\)