How do you find #dy/dx# by implicit differentiation of #y^2=(x^2-4)/(x^2+4)# and evaluate at point (2,0)?

Question

Evelyn Agard · Accepted Answer

#dy/dx# does not exist at #y=0#. Finding #dy/dx takes a bit more work.

Beginning with #y^2=(x^2-4)/(x^2+4)# we differentiate to get #2y dy/dx = "some function of "x#. But then #dy/dx# requires dividing by #y#, so it is not defined at #y = 0# In detail: #2ydy/dx = (2x(x^2+4)-2x(x^2-4))/(x^2+4)^2# #2ydy/dx = (16x)/(x^2+4)^2# #dy/dx = (8x)/(y(x^2+4)^2)# Here is the graph of the equation. graph{y^2-(x^2-4)/(x^2+4)=0 [-8.077, 7.724, -4.394, 3.51]}

Evelyn Agard · Answer

undefined To find $ \frac{dy}{dx} $ by implicit differentiation, differentiate both sides of the equation with respect to $ x $. Then solve for $ \frac{dy}{dx} $.

Given $ y^2 = \frac{x^2 - 4}{x^2 + 4} $, differentiate both sides with respect to $ x $ using implicit differentiation.

$$ \frac{d}{dx}(y^2) = \frac{d}{dx}\left(\frac{x^2 - 4}{x^2 + 4}\right) $$

$$ 2y \frac{dy}{dx} = \frac{(x^2 + 4) \frac{d}{dx}(x^2 - 4) - (x^2 - 4) \frac{d}{dx}(x^2 + 4)}{(x^2 + 4)^2} $$

$$ 2y \frac{dy}{dx} = \frac{(x^2 + 4)(2x) - (x^2 - 4)(2x)}{(x^2 + 4)^2} $$

$$ 2y \frac{dy}{dx} = \frac{2x(x^2 + 4) - 2x(x^2 - 4)}{(x^2 + 4)^2} $$

$$ 2y \frac{dy}{dx} = \frac{2x^3 + 8x - 2x^3 + 8x}{(x^2 + 4)^2} $$

$$ 2y \frac{dy}{dx} = \frac{16x}{(x^2 + 4)^2} $$

$$ \frac{dy}{dx} = \frac{8x}{y(x^2 + 4)^2} $$

Now, to evaluate at the point $ (2,0) $:

$$ x = 2 $$
$$ y = 0 $$

$$ \frac{dy}{dx} = \frac{8(2)}{0(2^2 + 4)^2} $$

$$ \frac{dy}{dx} = \frac{16}{0} $$

The derivative is undefined at $ (2,0) $.