How do you find an equation of the tangent line to the curve y=arcsin⁡(x2) at the point where x=−2?

Explanation:Let f(x)=arcsin⁡(x2). By the Chain Rule,f′(x)=11−(x2)2⋅12=14−x2Thus,f(−2)=arcsin⁡(−22)=−π4 andf′(−2)=14−2=12The equation of the tangent line isy=f(−2)+f′(−2)⋅(x−(−2))=−π4+12(x+2).This simplifies to: y=x2+1−π4

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High School
Calculus and Analysis
Calculus 1
Derivatives

Fastage6bk

Answered question

2022-02-12

How do you find an equation of the tangent line to the curve

y = \arcsin (\frac{x}{2})

at the point where

x = - \sqrt{2}

Answer & Explanation

lapinar7g1

Beginner2022-02-13Added 13 answers

Explanation:
Let

f (x) = \arcsin (\frac{x}{2})

. By the Chain Rule,

f^{'} (x) = \frac{1}{\sqrt{1 - {(\frac{x}{2})}^{2}}} \cdot \frac{1}{2} = \frac{1}{\sqrt{4 - x^{2}}}

Thus,

f (- \sqrt{2}) = \arcsin (- \frac{\sqrt{2}}{2}) = - \frac{π}{4}

and

f^{'} (- \sqrt{2}) = \frac{1}{\sqrt{4 - 2}} = \frac{1}{\sqrt{2}}

The equation of the tangent line is

y = f (- \sqrt{2}) + f^{'} (- \sqrt{2}) \cdot (x - (- \sqrt{2}))

= - \frac{π}{4} + \frac{1}{\sqrt{2}} (x + \sqrt{2}) .

This simplifies to:

y = \frac{x}{\sqrt{2}} + 1 - \frac{π}{4}

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