Implicit Differentiation Calculator with Steps

The implicit differentiation calculator will find the first and second derivatives of an implicit function treating either $y$ as a function of $x$ or $x$ as a function of $y$, with steps shown.

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Find $\frac{d}{dx} \left(x^{3} + y^{3} = 2 x y\right)$.

Solution

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

Differentiate the LHS of the equation.

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

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

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 3$:

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

The function $y^{3}{\left(x \right)}$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = u^{3}$ and $g{\left(x \right)} = y{\left(x \right)}$.

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

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

Apply the power rule $\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$ with $n = 3$:

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

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

Simplify:

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

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

Differentiate the RHS of the equation.

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 $c = 2$ and $f{\left(x \right)} = x y{\left(x \right)}$:

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

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 $g{\left(x \right)} = y{\left(x \right)}$:

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

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

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

Simplify:

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

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

Therefore, we have obtained the following linear equation with respect to the derivative: $3 x^{2} + 3 y^{2} \frac{dy}{dx} = 2 x \frac{dy}{dx} + 2 y$.

Solving it, we obtain that $\frac{dy}{dx} = \frac{3 x^{2} - 2 y}{2 x - 3 y^{2}}$.

$\frac{dy}{dx} = \frac{3 x^{2} - 2 y}{2 x - 3 y^{2}}$A