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Chapter 5: Problem 40

Find the derivative of the function. \(f(t)=\arcsin t^{2}\)

Short Answer

Expert verified
The derivative of the function \( f(t)=\arcsin t^{2} \) is \( \frac{2t}{\sqrt{1-t^{4}}} \)

Step by step solution

01

Identify the outer and inner functions

In this task, the function \( f(t) = \arcsin t^{2} \) can be identified as a composition of two functions. The outer function is \( \arcsin(x) \) and the inner function is \( t^{2} \). So, we can write \( f(t) = \arcsin(u) \) where \( u=t^{2} \) is the inner function.
02

Differentiation using the chain rule

We can use the chain rule, which states if a variable \( y \) is dependent on \( u \) and \( u \) in turn is dependent on \( t \), then the derivative of \( y \) with respect to \( t \) can be obtained as the derivative of \( y \) with respect to \( u \) and the derivative of \( u \) with respect to \( t \). In mathematical form, \( \frac{dy}{dt} = \frac{dy}{du} . \frac{du}{dt} \)
03

Calculate \( \frac{dy}{du} \) and \( \frac{du}{dt} \)

The derivative of \( u=t^{2} \) with respect to \( t \) where \( \frac{du}{dt}=2t \), we also know that the derivative of \( y=\arcsin(u) \) with respect to \( u \) is \( \frac{dy}{du} = \frac{1}{\sqrt{1-u^{2}}} \). We substitute \( u \( with \( t^{2} \) to get \( \frac{dy}{du} = \frac{1}{\sqrt{1-t^{4}}} \).
04

Substitute \( \frac{dy}{du} \) and \( \frac{du}{dt} \) into chain rule

Finally, we substitute \( \frac{dy}{du} \) and \( \frac{du}{dt} \) into the chain rule to find \( \frac{dy}{dt} \). After substitution we get \( \frac{dy}{dt} = \frac{1}{\sqrt{1-t^{4}}} . 2t = \frac{2t}{\sqrt{1-t^{4}}} \)

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