Question

find the derivative of y with respect to t.

( y = arcsin ( sqrt { 13 } t ) )

( \frac { d y } { d t } = square )
(simplify your answer. type an exact answer, using radicals as needed.)

Explanation:

Step1: Apply the chain rule

The chain rule states that if \(y = f(g(t))\), then \(\frac{dy}{dt}=f^{\prime}(g(t))\cdot g^{\prime}(t)\). For \(y=\arcsin(u)\) where \(u = \sqrt{13}t\), the derivative of \(\arcsin(u)\) with respect to \(u\) is \(\frac{1}{\sqrt{1 - u^{2}}}\), and the derivative of \(u=\sqrt{13}t\) with respect to \(t\) is \(\sqrt{13}\).

Step2: Substitute \(u\) and simplify

Substitute \(u = \sqrt{13}t\) into the formula. We get \(\frac{dy}{dt}=\frac{\sqrt{13}}{\sqrt{1-(\sqrt{13}t)^{2}}}\). Simplify the denominator: \(1 - 13t^{2}\). So \(\frac{dy}{dt}=\frac{\sqrt{13}}{\sqrt{1 - 13t^{2}}}\).

Answer:

\(\frac{\sqrt{13}}{\sqrt{1 - 13t^{2}}}\)