What is the equation of the tangent line of #f(x) =arcsin(cosx)# at #x=pi/4#?

Answer 1

#y=-x+pi/2#

Know that: #d/dx[arcsin(u)]=(u')/(sqrt(1-u^2))#

Thus:

#d/dx[arcsin(cosx)]=(d/dx[cosx])/(sqrt(1-cos^2x))#
#=(-sinx)/sqrt(sin^2x)=-sinx/abs(sinx)#
To find the slope of the tangent line when #x=pi/4#, find #f'(pi/4)#.
#f'(pi/4)=-sin(pi/4)/abs(sin(pi/4))=-(sqrt2/2)/(sqrt2/2)=-1#
The #y#-coordinate point that the tangent line will touch is:
#f(pi/4)=arcsin(cos(pi/4))=arcsin(sqrt2/2)=pi/4#
#(pi/4,pi/4)#

Equation should be written in point-slope form:

#y-pi/4=-(x-pi/4)#

Forming a point-intercept:

#y=-x+pi/2#

This graph's appearance is very interesting, despite its lack of relevance, and it demonstrates why the absolute value signs were required.

The function cannot be graphed by the built-in Socratic grapher, but you can try it out yourself by going to https://tutor.hix.ai

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Answer 2

The equation of the tangent line of f(x) = arcsin(cosx) at x = pi/4 is y = -√2x + √2 + pi/4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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