2. Ha#TH y' Hy'' 2.26. ) x=sqrt [3]((t-1)^2) y=sqrt (t-1)

Name: 2. HaTH Y' Hy'' 2.26. ) X=sqrt [3]((t-1)^2) Y=sqrt (t-1)
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Question

Answer 1

To find the derivatives of

and

with respect to

, we need to differentiate each equation separately.Given:

First, let's find

:

Using the chain rule:

Next, let's find

:

Using the chain rule:

So, the derivatives are:

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