How do you find an equation of the tangent line to the curve #y = arcsin(x/2)# at the point where #x = −sqrt2#?

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Answer 1 · 2015-06-26T14:19:26

#y=x/sqrt(2)+1-pi/4#

Let #f(x)=arcsin(x/2)#. By the Chain Rule,

#f'(x)=1/sqrt(1-(x/2)^2)*1/2=1/sqrt{4-x^2}#

Thus,

#f(-sqrt(2))=arcsin(-sqrt(2)/2)=-\pi/4# and

#f'(-sqrt(2))=1/sqrt{4-2}=1/sqrt(2)#

The equation of the tangent line is

#y=f(-sqrt(2))+f'(-sqrt(2))*(x-(-sqrt(2)))#

#=-pi/4+1/sqrt(2)(x+sqrt(2))#.

This simplifies to: #y=x/sqrt(2)+1-pi/4#