Implicit derivative of $x^{2} - x y + 3 y^{2} = 0$ with respect to $x$

The calculator will find the first and second derivatives of the implicit function $x^{2} - x y + 3 y^{2} = 0$ with respect to $x$, with steps shown.
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Find $\frac{d}{dx} \left(x^{2} - x y + 3 y^{2} = 0\right)$.

Solution

Differentiate separately both sides of the equation (treat $y$ as a function of $x$): $\frac{d}{dx} \left(x^{2} - x y{\left(x \right)} + 3 y^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(0\right)$.

Differentiate the LHS of the equation.

The derivative of a sum/difference is the sum/difference of derivatives:

$${\color{red}\left(\frac{d}{dx} \left(x^{2} - x y{\left(x \right)} + 3 y^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right) + \frac{d}{dx} \left(3 y^{2}{\left(x \right)}\right)\right)}$$

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 2$:

$${\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} - \frac{d}{dx} \left(x y{\left(x \right)}\right) + \frac{d}{dx} \left(3 y^{2}{\left(x \right)}\right) = {\color{red}\left(2 x\right)} - \frac{d}{dx} \left(x y{\left(x \right)}\right) + \frac{d}{dx} \left(3 y^{2}{\left(x \right)}\right)$$

Apply the constant multiple rule $\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$ with $c = 3$ and $f{\left(x \right)} = y^{2}{\left(x \right)}$:

$$2 x + {\color{red}\left(\frac{d}{dx} \left(3 y^{2}{\left(x \right)}\right)\right)} - \frac{d}{dx} \left(x y{\left(x \right)}\right) = 2 x + {\color{red}\left(3 \frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)} - \frac{d}{dx} \left(x y{\left(x \right)}\right)$$

The function $y^{2}{\left(x \right)}$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = u^{2}$ and $g{\left(x \right)} = y{\left(x \right)}$.

Apply the chain rule $\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$:

$$2 x + 3 {\color{red}\left(\frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)} - \frac{d}{dx} \left(x y{\left(x \right)}\right) = 2 x + 3 {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} - \frac{d}{dx} \left(x y{\left(x \right)}\right)$$

Apply the power rule $\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$ with $n = 2$:

$$2 x + 3 {\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right) = 2 x + 3 {\color{red}\left(2 u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right)$$

$$2 x + 6 {\color{red}\left(u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right) = 2 x + 6 {\color{red}\left(y{\left(x \right)}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right)$$

Apply the product rule $\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$ with $f{\left(x \right)} = x$ and $g{\left(x \right)} = y{\left(x \right)}$:

$$2 x + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - {\color{red}\left(\frac{d}{dx} \left(x y{\left(x \right)}\right)\right)} = 2 x + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - {\color{red}\left(\frac{d}{dx} \left(x\right) y{\left(x \right)} + x \frac{d}{dx} \left(y{\left(x \right)}\right)\right)}$$

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dx} \left(x\right) = 1$:

$$- x \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x - y{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) = - x \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x - y{\left(x \right)} {\color{red}\left(1\right)} + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right)$$

Thus, $\frac{d}{dx} \left(x^{2} - x y{\left(x \right)} + 3 y^{2}{\left(x \right)}\right) = - x \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - y{\left(x \right)}.$

Differentiate the RHS of the equation.

The derivative of a constant is $0$:

$${\color{red}\left(\frac{d}{dx} \left(0\right)\right)} = {\color{red}\left(0\right)}$$

Thus, $\frac{d}{dx} \left(0\right) = 0$.

Therefore, we have obtained the following linear equation with respect to the derivative: $- x \frac{dy}{dx} + 2 x + 6 y \frac{dy}{dx} - y = 0$.

Solving it, we obtain that $\frac{dy}{dx} = \frac{2 x - y}{x - 6 y}$.

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