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9-12.

Multiple Choice: The equation of the line tangent to $y = \tan^{–1}(x)$ at the point where $x =\sqrt { 3 }$ is:

$y =\frac { 1 } { 4 }x -\frac { \sqrt { 3 } } { 4 }+\frac { \pi } { 3 }$

$y =\frac { 1 } { 4 }x-\frac { \sqrt { 3 } } { 4 }+\frac { \pi } { 6 }$

$y = −\frac { 1 } { 2 }x+\frac { \sqrt { 3 } } { 2 }+\frac { \pi } { 3 }$

$y = −\frac { 1 } { 2 }x+\frac { \sqrt { 3 } } { 2 }+\frac { \pi } { 6 }$

Step 1:

$y'=\frac{1}{1+x^2}$

Step 2:

$y'(\sqrt{3})=\frac{1}{4}$

Step 3:

The point of tangency is:

$(\sqrt{3},\tan^{-1}(\sqrt{3}))$

Step 4:

Use the slope and the point of tangency to write the equation of the line.

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