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.

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

Differentiate with respect to:

Evaluate at $$$\left(x_{0}, y_{0}\right)$$$:$$$($$$

,

$$$)$$$

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Your Input

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:

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

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

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

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

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

Return to the old variable:

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

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

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

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

Answer

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