Chapter 2: Q. 12 (page 232)

The function $\sin^{- 1} x$ is defined on $[- 1, 1]$ , but its derivative $\frac{1}{\sqrt{1 - x^{2}}}$ is defined only on $(- 1, 1)$ . Explain why the tangent lines to the graph of $y = \sin^{- 1} x$ do not exist at $x = \pm 1 .$ (Hint: Think about the corresponding tangent lines on the graph of the restricted sine function.)

Short Answer

Expert verified

The derivative does not exist at $x = \pm 1,$ therefore the tangent lines does not exist at $x = \pm 1 .$

Step by step solution

Step 1. Given Information.

The given function is $y = \sin^{- 1} x$ .

The derivative is $\frac{1}{\sqrt{1 - x^{2}}} .$

Step 2. Explanation.

$x = \pm 1 .$ Let $f^{'} (x) = \frac{d}{d x} \sin^{- 1} x = \frac{1}{\sqrt{1 - x^{2}}}$

We know, slope is given by,

$f^{'} (x)$

Therefore,

$f^{'} (- 1) = \frac{1}{\sqrt{1 - (- 1)^{2}}} = \frac{1}{0} and f^{'} (1) = \frac{1}{\sqrt{1 - (1)^{2}}} = \frac{1}{0}$

That is not defined.

Hence, derivative does and exist and therefore, tangent lines do not exist at $x = \pm 1 .$

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The function sin-1xis defined on [-1,1], but its derivative 11-x2is defined only on (-1,1). Explain why the tangent lines to the graph of y=sin-1x do not exist at x=±1. (Hint: Think about the corresponding tangent lines on the graph of the restricted sine function.)

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Explanation.

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