How do you implicitly differentiate # xy+2x+3x^2=-4#?

Question

How do you implicitly differentiate #  xy+2x+3x^2=-4#?

Isaiah Allen · Accepted Answer

So, recall that for implicit differentiation, each term has to be differentiated with respect to a single variable, and that to differentiate some #f(y)# with respect to #x#, we utilise the chain rule: #d/dx(f(y)) = f'(y)*dy/dx# Thus, we state the equality:
#d/dx(xy) + d/dx(2x) + d/dx(3x^2) = d/dx(-4)#
#rArr x*dy/dx + y + 2 +6x = 0# (using the product rule to differentiate #xy#). Now we just need to sort out this mess to get an equation #dy/dx =...# #x*dy/dx = -6x-2-y#
#:. dy/dx = -(6x+2+y)/x# for all #x in RR# except zero.

Emily Ammerman · Answer

undefined To implicitly differentiate the equation $xy + 2x + 3x^2 = -4$, follow these steps:

1. Differentiate each term with respect to $x$.
2. Use the product rule for the term $xy$.
3. Apply the power rule for the terms $2x$ and $3x^2$.
4. Simplify the resulting expression.

The implicit differentiation of $xy + 2x + 3x^2 = -4$ with respect to $x$ yields:

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

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

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

$$ x\frac{dy}{dx} = -(y + 2 + 6x) $$

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