Given #ye^x - x - y^2 = 0# calculate #dy/dx# ?

Question

Given #ye^x  - x - y^2 = 0# calculate #dy/dx# ?

Ezra Adams · Accepted Answer

#(dy)/(dx) = ( 1-e^x y)/(e^x - 2 y)# Calling #f(x,y) = ye^x - x - y^2 = 0# we have #(dy)/(dx) = - f_x/(f_y) = -( e^x y-1)/(e^x - 2 y)#

Elijah Abram · Answer

undefined To find $ \frac{dy}{dx} $ from the equation $ ye^x - x - y^2 = 0 $, we'll use implicit differentiation:

$$ \frac{d}{dx}(ye^x) - \frac{d}{dx}(x) - \frac{d}{dx}(y^2) = 0 $$

$$ ye^x + y\frac{d}{dx}(e^x) - 1 - 2y\frac{dy}{dx} = 0 $$

$$ ye^x + ye^x - 2y\frac{dy}{dx} - 1 = 0 $$

$$ 2ye^x - 2y\frac{dy}{dx} = 1 $$

$$ \frac{dy}{dx} = \frac{2ye^x - 1}{2y} $$

Cameron Alexander · Answer

undefined To find $ \frac{dy}{dx} $ from the equation $ ye^x - x - y^2 = 0 $, you can use implicit differentiation. First, differentiate each term with respect to $ x $ using the product rule and chain rule where applicable. Then, solve for $ \frac{dy}{dx} $.

Differentiating $ ye^x - x - y^2 = 0 $ with respect to $ x $, we get:

$$ \frac{d}{dx}(ye^x) - \frac{d}{dx}(x) - \frac{d}{dx}(y^2) = 0 $$

$$ y\frac{d}{dx}(e^x) + e^x\frac{dy}{dx} - 1 - 2y\frac{dy}{dx} = 0 $$

$$ ye^x + e^x\frac{dy}{dx} - 1 - 2y\frac{dy}{dx} = 0 $$

Now, rearranging terms and solving for $ \frac{dy}{dx} $, we get:

$$ e^x\frac{dy}{dx} - 2y\frac{dy}{dx} = 1 - ye^x $$

$$ \frac{dy}{dx}(e^x - 2y) = 1 - ye^x $$

$$ \frac{dy}{dx} = \frac{1 - ye^x}{e^x - 2y} $$

So, $ \frac{dy}{dx} = \frac{1 - ye^x}{e^x - 2y} $.