Question

find the differential of $y = sqrt{10 + t^{2}}$.
$dy=square dt$

Explanation:

Step1: Rewrite the function

Let $y=(10 + t^{2})^{\frac{1}{2}}$.

Step2: Apply chain - rule

The chain - rule states that if $y = f(u)$ and $u = g(t)$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. Let $u = 10 + t^{2}$, so $y = u^{\frac{1}{2}}$. First, find $\frac{dy}{du}$ and $\frac{du}{dt}$.
$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dt}=2t$.

Step3: Calculate $\frac{dy}{dt}$

$\frac{dy}{dt}=\frac{1}{2}(10 + t^{2})^{-\frac{1}{2}}\cdot2t=\frac{t}{\sqrt{10 + t^{2}}}$.

Step4: Find the differential

Since $dy=\frac{dy}{dt}dt$, then $dy=\frac{t}{\sqrt{10 + t^{2}}}dt$.

Answer:

$\frac{t}{\sqrt{10 + t^{2}}}$