**Implicit Differentiation**

Because this question was posted under Implicit Differentiation, we'll do it that way.

We want #dy/dx# given that #x# and #y# are related by:

#x^2 + y^2 = 12#

Differentiate both sides with respect to #x#. Remember that we are considering #y# to be some function of #x# that we haven't found. It often helps to think of #y# as "some stuff in parentheses".

When we differentiate "some stuff in parentheses" squared, we gat #2# times that stuff **times the derivative of the stuff** (because of the chain rule).

Also: remember that #d/dx# of some expression means the derivative (with respect to x) of that expression.

o, (in moderate detail)

#x^2 + y^2 = 12#

#d/dx(x^2 + y^2) = d/dx(12)#

#d/dx(x^2) + d/dx(y^2) = d/dx(12)#

#2x + underbrace(2y dy/dx)_"used chain rule" = 0#

#dy/dx = -x/y#

At #(sqrt3,3)#, the slope is #m = -sqrt3/3#

The equation of the line through that point with that slope is:

#y=-sqrt3/3x+4#

**Notes**

**(1)** because we see that the #y# coordinate of the point of interest is positive, we know that the point lies on the upper half of the circle:

#x^2 + y^2 = 12#

Solving for #y# on that semicircle gets us #y = sqrt(12-x^2)# and we do not need implicit differentiation.

**(2)** It is a circle centered at the origin, The tangent at #(sqrt3,3)# is perpendicular to the radius connecting #(0,0)# and #(sqrt3, 3)#. Find the slope of the radius and then the slope of a perpendicular to the radius. No need for calculus.