How do you find #(dy)/(dx)# given #x^2+y^2=1#?

Question

Ezekiel Alexander · Accepted Answer

# dy/dx=-x/y # #x^2+y^2=1# Differentiate wrt #x#: # d/dxx^2+d/dxy^2=d/dx1 # We already know how to deal with the first and third terms, so lets get them out the way: # d/dxx^2+d/dxy^2=d/dx1 # # :.2x+d/dxy^2=0 # For the remaining term we use the chain rule, we don't know how to differentiate #y^2# wrt #x# but we do know how to differentiate #y^2# wrt #y# (it the same as differentiating #x^2# wrt #x#!). The chain rule tells us that: #dy/dx=dy/(du)*(du)/dx#, so we can rewrite: # 2x+d/dxy^2=0 # as # 2x+d/dy(y^2)*dy/dx=0 # We can know perform that final differentiation, as we are now differentiating a function of #y# wrt #y# so # 2x+d/dy(y^2)*dy/dx=0 # # :. 2x+2y*dy/dx=0 # We can then rearrange to get #dy/dx# as follows: # 2x+2y*dy/dx=0 # # :. 2y*dy/dx=-2x # # :. dy/dx=-(2x)/(2y) # # :. dy/dx=-x/y #

Isaiah Allen · Answer

undefined To find (dy)/(dx) given x^2+y^2=1, you can implicitly differentiate both sides of the equation with respect to x, and then solve for (dy)/(dx).

Differentiating both sides of the equation x^2 + y^2 = 1 with respect to x:

d/dx(x^2) + d/dx(y^2) = d/dx(1)

2x + 2y(dy/dx) = 0

Now, solve for (dy)/(dx):

2y(dy/dx) = -2x

(dy/dx) = (-2x) / (2y)

(dy/dx) = -x / y