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Chapter 5: Problem 49

In Exercises \(41-56,\) 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

Apply the chain rule

To differentiate \(h(t) = \sin(\arccos t)\), the chain rule is used. According to the chain rule, the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In this case, the outer function is \(\sin(x)\) and the inner function is \(\arccos t\). The derivative of \(\sin(x)\) is \(\cos(x)\) and the derivative of \(\arccos(t)\) is \(-1/\sqrt{1-t^2}\). Therefore, applying the chain rule gives \(h'(t) = \cos(\arccos t) \cdot -1/\sqrt{1-t^2}\).
02

Simplify Using Trigonometric Identity

The derivative at this stage is still not entirely simplified. One can use the trigonometric identity to further simplify it. The fact that \(\cos(\arccos(t))\) equals \(t\) is useful. Thus, the derivative \(h'(t) = \cos(\arccos t) \cdot -1/\sqrt{1-t^2}\) simplifies to \(h'(t) = -t/\sqrt{1-t^2}.\)
03

Final derivative of the function

After performing these steps, the derivative of the function \(h(t) = \sin(\arccos t)\) is found to be \(h'(t) = -t/\sqrt{1-t^2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. When you have a function inside another function, like in our example with \( h(t) = \sin(\arccos t) \), the chain rule helps break this complex differentiation into manageable steps.
The rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
  • Identify the outer function: In \(h(t)\), it's \(\sin\), whose derivative is \(\cos\).
  • Identify the inner function: Here, it's \(\arccos t\), which differentiates to \(-1/\sqrt{1-t^2}\).

Combining these, you get \(h'(t) = \cos(\arccos t) \times \left(-\frac{1}{\sqrt{1-t^2}}\right)\). Using the chain rule streamlines the differentiation process for nested functions.
Trigonometric Identity
Trigonometric identities are true equations that involve trigonometric functions. One such identity plays a crucial role in our derivative simplification process.
In step 2 of the solution, the derivative is initially \( h'(t)= \cos(\arccos t) \cdot -1/\sqrt{1-t^2} \).
Here, we utilize the identity \( \cos(\arccos(t)) = t \) to make the expression simpler.
  • \( \arccos(t) \) gives you an angle whose cosine value is \(t\).
  • You apply this identity to directly replace \( \cos(\arccos t) \) with \(t\) in the derivative.

So, the derivative expression transforms into \(-t/\sqrt{1-t^2}\). Trigonometric identities make complex trigonometric expressions much easier to handle, as they allow you to recognize patterns and simplify expressions accordingly.
Composite Function
A composite function, essentially a function within another function, requires a special approach for differentiation. In this exercise, \( h(t) = \sin(\arccos t) \) is a classic example of a composite function, combining \(\sin\) and \(\arccos\) into one.
Understanding the structure of a composite function is vital as it dictates the application of the chain rule during differentiation.
  • The '**outer function**' is the function that wraps around another, in our case, it is \( \sin(x) \).
  • The '**inner function**' is the one nested inside, which is \( \arccos t \) here.

By identifying these components, you can correctly apply mathematical operations, like the chain rule, to differentiate or simplify the function. Handling composite functions might initially seem tricky, but they frequently appear in calculus, making it essential to become comfortable with them.

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Most popular questions from this chapter

Finding an Equation of a Tangent Line In Exercises \(61-64,\) find an equation of the tangent line to the graph of the function at the given point. $$y=\log _{3} x, \quad(27,3)$$

Finding an Equation of a Tangent Line In Exercises \(67-74,\) (a) find an equation of the tangent line to the graph of the function at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the tangent feature of a graphing utility to confirm your results. $$f(x)=\frac{1}{2} x \ln x^{2}, \quad(-1,0)$$

In Exercises 51 and 52, show that the antiderivatives are equivalent. $$\int \frac{6}{4+9 x^{2}} d x=\arctan \frac{3 x}{2}+C \text { or arccsc } \frac{\sqrt{4+9 x^{2}}}{3 x}+C$$

Comparing Rates of Growth Order the functions $$f(x)=\log _{2} x, g(x)=x^{x}, h(x)=x^{2}, and k(x)=2^{x}$$ from the one with the greatest rate of growth to the one with the least rate of growth for large values of \(x .\)

Using the Area of a Region Find the value of \(a\) such that the area bounded by \(y=e^{-x},\) the \(x\) -axis, \(x=-a,\) and \(x=a\) is \(\frac{8}{3}\)
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