How do you implicitly differentiate # 2x+2y = sqrt(x^2+y^2)#?

Question

How do you implicitly differentiate #  2x+2y = sqrt(x^2+y^2)#?

Emma Aldridge · Accepted Answer

#dy/dx = { -1/3 ( sqrt[7]+4), 1/3 ( sqrt[7]-4)}#

depending on de sign of #x# #f(x,y)=2x+2y - sqrt(x^2+y^2)=0# is an homogeneus function. Making the substitution #y = lambda x# we get at #f(x,lambda x) = 2x(1+lambda)-abs(x) sqrt(1+lambda^2) = 0# or #x/abs(x)=1/2sqrt(1+lambda^2)/(1+lambda) = pm 1# Solving for #lambda# we have #x/abs(x) = 1-> lambda = 1/3 ( sqrt[7]-4)# #x/abs(x) = -1-> lambda = -1/3 ( sqrt[7]+4)# so #dy/dx = { -1/3 ( sqrt[7]+4), 1/3 ( sqrt[7]-4)}# depending on de sign of #x#

Christopher Allers · Answer

undefined To implicitly differentiate the equation $2x + 2y = \sqrt{x^2 + y^2}$, we'll apply the differentiation rules step by step. Let's begin:

Given:
$$2x + 2y = \sqrt{x^2 + y^2}$$

Differentiating both sides with respect to $x$ gives us:

$$2 + 2\frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + y^2}}(2x + 2y\frac{dy}{dx})$$

This stems from applying the chain rule to the right-hand side, recognizing that $y$ is a function of $x$, hence $\frac{d}{dx}y = \frac{dy}{dx}$.

To solve for $\frac{dy}{dx}$, we'll manipulate the equation:

$$2 + 2\frac{dy}{dx} = \frac{x + y\frac{dy}{dx}}{\sqrt{x^2 + y^2}}$$

Multiplying both sides by $\sqrt{x^2 + y^2}$ to clear the fraction:

$$2\sqrt{x^2 + y^2} + 2\frac{dy}{dx}\sqrt{x^2 + y^2} = x + y\frac{dy}{dx}$$

Rearrange to collect $\frac{dy}{dx}$ terms on one side:

$$2\frac{dy}{dx}\sqrt{x^2 + y^2} - y\frac{dy}{dx} = x - 2\sqrt{x^2 + y^2}$$

Factor out $\frac{dy}{dx}$:

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

Finally, solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{x - 2\sqrt{x^2 + y^2}}{2\sqrt{x^2 + y^2} - y}$$

This gives the derivative of $y$ with respect to $x$ implicitly as a function of both $x$ and $y$.