Find the equation of the tangent line to the curve x^2 + y^2 = (2x^2 + 2y^2 -x)^2 at the point 0, 1/2?

#x^2+y^2 = (2x^2 + 2y^2 - x)^2#

Differentiating term by term w.r.t. x

That means simple x terms differentiate normally but while differentiating those with y; since you are differentiating with x; you'll have to multiply those with #dy/dx#.

Step by step differentiation:

#x^2+y^2 = (2x^2 + 2y^2 - x)^2#

#2x+2y (dy/dx) = 2 (2x^2 + 2y^2 -x)(4x + 4y(dy/dx) - 1)#

# x + y(dy/dx) = (2x^2 + 2y^2 - x)(4x + 4y(dy/dx) - 1)#

I hope you understood what I did on the RHS of the second to last step; I treated the whole term first as a huge x term that I had to differentiate and then following the chain rule I differentiated the inside and multiplied it.

Next, expanding and separating out the #dy/dx# terms to one side:
#x+y(dy/dx) = 8x^3 + 8x^2y(dy/dx) - 2x^2 +8xy^2 + 8y^3(dy/dx) - 2y^2 - 4x^2 - 4xy(dy/dx) +x#

#dy/dx = (8x^3 - 6x^2 +8xy^2 - 2y^2)/(y-8x^2y+8y^3-4xy)#

Equation of tangent to line is given by:

#y-y_1 = (dy/dx)_(x_1,y_1) (x-x_1)#

where #(x_1,y_1) = (0,1/2)#

Putting values of #(x_1,y_1)# in the equation we got for #dy/dx#:

#(dy/dx)_(x_1,y_1) = -1/3#

Putting it into the equation of tangent:

#y - 1/2 = (-1/3)(x-0)#

Simplifying:

# 2x + 6y - 3 = 0#

And there you have it! It was a pretty simple question. Hope it helped!

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

#2x+2y (dy/dx) = 2 (2x^2 + 2y^2 -x)(4x + 4y(dy/dx) - 1)#

We were not asked for a general formula for #dy/dx# only for its value at #(0,1/2)#. So let's plug in the numbers and solve for #dy/dx#

#x = (0)# and #y = (1/2)#

#2(0)+2(1/2)(dy/dx) = 2 (2(0)^2 + 2(1/2)^2 -(0))(4(0) + 4(1/2)(dy/dx) - 1)#

#dy/dx = 2(1/2)(2dy/dx-1)#

#dy/dx = 2dy/dx-1#

#dy/dx = 1#

So the tangent line has slope #@1# and #y#-intercept #1/2#.

#y = x+1/2#