How do you implicitly differentiate #x+xy-2x^3 = 2#?

Question

Emma Aldridge · Accepted Answer

Let us define #f(x,y) = x + xy - 2x^3 = 2#. #(df)/(dx) = (df)/(dy)*(dy)/(dx)#, and so every time you differentiate (#(df)/(dy)#) a part of the function that has #y#, you have to multiply by #(dy)/(dx)#, the derivative of #y# with respect to #x#, for the overall derivative (#(df)/(dx)#) to still be with respect to #x# (even though you're differentiating #y#). #(df)/(dx) = (d(x))/(dx) + [x*(dy)/(dx) + y*(d(x))/(dx)] - (d(2x^3))/(dx) = cancel((d(0))/(dx))#
(has product rule) #1 + [x(dy)/(dx) + y] - 6x^2 = 0# Simple algebra from here:
#x(dy)/(dx)= 6x^2 - 1 - y# #(dy)/(dx) = (6x^2 - y - 1)/x#

Ezekiel Alexander · Answer

undefined To implicitly differentiate $x + xy - 2x^3 = 2$, follow these steps:

1. Differentiate each term with respect to $x$.
2. Apply the chain rule whenever differentiating terms involving $y$.
3. Collect terms involving $y'$ on one side of the equation.
4. Solve for $y'$.

The steps are as follows:

1. Differentiate each term:

$$ \frac{d}{dx}(x) + \frac{d}{dx}(xy) - \frac{d}{dx}(2x^3) = \frac{d}{dx}(2) $$

$$ 1 + x \frac{dy}{dx} + y - 6x^2 = 0 $$

2. Rearrange terms:

$$ x \frac{dy}{dx} + y = -1 + 6x^2 $$

3. Solve for $y'$:

$$ \frac{dy}{dx} = \frac{-1 + 6x^2 - y}{x} $$