# Find the slope of the tangent line to the curve at the given points in two ways: first by solving for y in terms of x and differentiating and then by implicit differentiation.
* x^2 + y^2 = 1; (1/2,√3/2), (1/2,−√3/2)

Question

# Find the slope of the tangent line to the curve at the given points in two ways: first by solving for y in terms of x and differentiating and then by implicit differentiation.
* x^2 + y^2 = 1; (1/2,√3/2), (1/2,−√3/2)

welshfella · Answer

have you done any work on this so far?

anonymous · Answer

x^2 + y^2 = 1
y²=1-x^²
y=$$\sqrt{1-x²}$$
y=(1-x²)$$^{1/2}$$
implicit method:
d/dx (x²+y²=1)
2x+2yy'=0
y'=-2x/2y=-x/y
 =1/2$$(1-x²)^{-1/2}$$(-2x)
 =$$-x/\sqrt{1-x²}$$

welshfella · Answer

to find the values of the slope plug the values of x into  the derivative

-x  /  sqrt(1 - x^2)