How do you differentiate #(y+3x)^2-4x=0#?

Question

Paisley Johnson · Accepted Answer

Assuming that we want #dy/dx#, we get: #2(y+3x)d/dx(y+3x) - 4 = 0# #2(y+3x)(dy/dx+3) - 4 = 0# Now solve for #dy/dx# #dy/dx = 2/(y+3x) -3# Rewrite using algebra until you like the way it looks.

Adam Adkins · Answer

undefined To differentiate (y + 3x)^2 - 4x = 0, you would apply the chain rule for differentiation. The chain rule states that if you have a function inside another function, you differentiate the outer function with respect to the inner function, then multiply by the derivative of the inner function.

So, differentiating with respect to x, you get:

$$ \frac{d}{dx}((y + 3x)^2 - 4x) = 2(y + 3x) \frac{d}{dx}(y + 3x) - 4 $$

$$ = 2(y + 3x) \frac{dy}{dx} + 6x - 4 $$

$$ = 2y\frac{dy}{dx} + 6x\frac{dy}{dx} + 6x - 4 $$

$$ = (2y + 6x)\frac{dy}{dx} + 6x - 4 $$

That's the derivative of (y + 3x)^2 - 4x with respect to x.

Emma Aldridge · Answer

undefined To differentiate the equation $(y + 3x)^2 - 4x = 0$ with respect to $x$, you can follow these steps:

1. Expand the expression $(y + 3x)^2$ using the binomial square formula.
2. Differentiate each term in the expanded expression with respect to $x$.
3. Set the resulting expression equal to zero and solve for $y'$, which represents the derivative of $y$ with respect to $x$.

Let's differentiate step by step:

1. Expand $(y + 3x)^2$:
$$ (y + 3x)^2 = y^2 + 6xy + 9x^2 $$

2. Differentiate each term with respect to $x$:
$$ \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} $$
$$ \frac{d}{dx}(6xy) = 6y + 6x \frac{dy}{dx} $$
$$ \frac{d}{dx}(9x^2) = 18x $$

3. Now, substitute the differentiated terms into the original equation:
$$ 2y \frac{dy}{dx} + 6y + 6x \frac{dy}{dx} + 18x - 4 = 0 $$

4. Combine like terms and solve for $y'$:
$$ 2y \frac{dy}{dx} + 6x \frac{dy}{dx} = -6y - 18x + 4 $$
$$ \frac{dy}{dx}(2y + 6x) = -6y - 18x + 4 $$
$$ \frac{dy}{dx} = \frac{-6y - 18x + 4}{2y + 6x} $$

So, the derivative of $y$ with respect to $x$ is $\frac{-6y - 18x + 4}{2y + 6x}$.