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Chapter 5: Problem 47

Find the derivative of the function. \(h(t)=\sin (\arccos t)\)

Short Answer

Expert verified
The derivative of the function \(h(t)=\sin (\arccos t)\) is \(h'(t) = -t/\sqrt{1-t^2}\)

Step by step solution

01

Identify the inner and outer function

In the given function, \(h(t)=\sin (\arccos t)\), the inside function is \(\arccos t\) and the outside function is \(\sin(x)\). This must be clearly understood before proceeding. The inner function has its domain in \([-1, 1]\).
02

Apply the Chain Rule

To find the derivative of \(h(t)\), use the chain rule. The chain rule states that the derivative of a composition of functions is the derivative of the outer function times the derivative of the inner function. So begin by taking the derivative of \(\sin(x)\) which is \(\cos(x)\) and use \(\arccos t\) as \(x\). Thus, the first part of the chain rule is \(\cos(\arccos t)\).
03

Complete the Chain Rule

Now, the second part of the chain rule is taking the derivative of the inner function, which is \(\arccos t\). The derivative of \(\arccos t\) is \(-1/\sqrt{1-t^2}\). Multiply this with the result from Step 2. Hence, the derivative \(h'(t)\) is \(\cos(\arccos t)\) * \((-1)/\sqrt{1-t^2}\).
04

Simplify the Result

Note that \(\cos(\arccos t)\) simplifies to \(t\). Therefore, plugging this into our derivative gives us: \(h'(t) = t*(-1)/\sqrt{1-t^2}\)
05

Final Result

Finally, a simplified version of the derivative is given by \(h'(t) = -t/\sqrt{1-t^2}\)

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