How do you find #dy/dx# by implicit differentiation given #e^y=x^2+y#?

Question

Luke Alvarez · Accepted Answer

Note that since #y = y(x)#, taking the derivative of the entire function with respect to #y# requires you to account for that fact using the chain rule. #(df)/(dx) = (df)/(dy)(dy)/(dx)# where #(df)/(dy)# is the derivative of, say, #e^y# or #y# in your given function. Thus: #e^y (dy)/(dx) = 2x + 1(dy)/(dx)# #=> (e^y - 1)(dy)/(dx) = 2x# #=> color(blue)((dy)/(dx) = (2x)/(e^y - 1))# If you know what #y# is in terms of #x#, you can plug it back in to give you a function containing just #x#.

Axel Alvarez · Answer

undefined To find $ \frac{dy}{dx} $ by implicit differentiation for the equation $ e^y = x^2 + y $, follow these steps:

1. Take the derivative of both sides of the equation with respect to $ x $.
2. Apply the chain rule for the derivative of $ e^y $.
3. Apply the sum rule for the derivative of $ x^2 + y $.
4. Isolate $ \frac{dy}{dx} $ on one side of the equation.

Here's the solution:

Starting with the given equation:

$$ e^y = x^2 + y $$

1. Take the derivative of both sides with respect to $ x $:

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

2. Apply the chain rule for the derivative of $ e^y $:

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

3. Apply the sum rule for the derivative of $ x^2 + y $:

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

4. Isolate $ \frac{dy}{dx} $ on one side:

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

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

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

So, $ \frac{dy}{dx} = \frac{2x}{e^y - 1} $ is the derivative of $ y $ with respect to $ x $.