How do find the gradient of the tangent for this function?

Question

#sqrt#1-#x^2# at #x#=#1/2#

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Answer 1 · 2018-02-26T17:16:46

Gradient at #x=1/2# is #-1/sqrt3#

Gradient or slope of function is the value of its derivative at that point.

As #y=sqrt(1-x^2)#

#(dy)/(dx)=1/(2sqrt(1-x^2))xx(-2x)=-x/sqrt(1-x^2)#

and at #x=1/2#, gradient is #-(1/2)/sqrt(1-(1/2)^2)#

= #-(1/2)/sqrt(3/4)#

= #-(1/2)/(sqrt3/2)#

= #-1/sqrt3#

graph{(y-sqrt(1-x^2))(y+x/sqrt3-1/(2sqrt3)-sqrt3/2)=0 [-2.469, 2.53, -0.2, 2.3]}