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Chapter 2: Q. 44 (page 233)

Find the derivatives of each of the functions in Exercises 17鈥�50. In some cases it may be convenient to do some preliminary algebra.
$f (x) = \sin^{2} (\sec^{1} x)$

Short Answer

Expert verified

The derivative is $\frac{2 \sin (\sec^{- 1} x) (\cos (\sec^{- 1} x))}{| x | \sqrt{x^{2} - 1}}$ .

Step by step solution

01

Step 1. Given Information.

The given function is $f (x) = \sin^{2} (\sec^{- 1} x)$ .

02

Step 2. Preliminary Algebra.

We know,

$C h a i n R u l e : (f 鈭� u)^{'} (x) = f^{'} (u (x)) u^{'} (x) A l s o, \frac{d}{d x} x^{n} = n x^{n - 1} \frac{d}{d x} (\sec^{- 1} x) = \frac{1}{| x | \sqrt{x^{2} - 1}} \frac{d}{d x} (\sin x) = \cos x$

03

Step 3. Derivative of the function.

The derivative of the function is ,

$\frac{d}{d x} (\sin^{2} (\sec^{- 1} x)) = \frac{d}{d x} {(\sin (\sec^{- 1} x))}^{2} = 2 \sin (\sec^{- 1} x) \frac{d}{d x} \sin (\sec^{- 1} x) = 2 \sin (\sec^{- 1} x) (\cos (\sec^{- 1} x)) \frac{d}{d x} (\sec^{- 1} x) = 2 \sin (\sec^{- 1} x) (\cos (\sec^{- 1} x)) \frac{1}{| x | \sqrt{x^{2} - 1}} = \frac{2 \sin (\sec^{- 1} x) (\cos (\sec^{- 1} x))}{| x | \sqrt{x^{2} - 1}}$

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Most popular questions from this chapter

A bowling ball dropped from a height of $400$ feet will be $s (t) = 400 - 16 t^{2}$ feet from the ground after $t$ seconds. Use a sequence of average velocities to estimate the instantaneous velocities described below:

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