Differentiating Explicit and Implicit Functions

An explicit function is one where the dependent variable is written directly in terms of the independent variable.

For example:

\[ y = x^2 + 3x - 8 \]

Here, y is the dependent variable and is expressed explicitly in terms of the independent variable x.

Note that y is the subject of the formula

An implicit function involves both variables mixed together in an equation.

For example:

\[ y + x^2 - 3x + 8 = 0 \]

Sometimes it is inconvenient or impossible to rearrange a function explicitly. For example, the circle:

\[ x^2 + y^2 = 16 \]

could be written explicitly as:

\[ y = \sqrt{16 - x^2} \qquad\text{or}\qquad y = -\sqrt{16 - x^2} \]

But which version should be used when differentiating?

It is often easier to differentiate an implicit function without having to rearrange it, by differentiating each term in turn.

Implicit Differentiation

Differentiate each term with respect to $x$.
Remember: because $y$ is a function of $x$, the chain , product and quotient rules apply!

Example

Differentiate $ x^2 + y^2 = 16 $ with respect to x.

Differentiate term‑by‑term:

\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0 \] \[ 2x + 2y\frac{dy}{dx} = 0 \]

Rearrange:

\[ \frac{dy}{dx} = -\frac{x}{y}. \]

This is far quicker than rearranging the circle into $y = \pm\sqrt{16 - x^2}$ and differentiating explicitly.

\[ x^{2} + y^{2} = 16 \] \[ \Rightarrow\; y^{2} = 16 - x^{2} \] \[ \Rightarrow\; y = \sqrt{\,16 - x^{2}\,} \] \[ \Rightarrow\; y = \left(16 - x^{2}\right)^{1/2} \] \[ \Rightarrow\; \frac{dy}{dx} = \frac{1}{2}\left(16 - x^{2}\right)^{-1/2}\cdot(-2x) \] \[ \Rightarrow\; \frac{dy}{dx} = -x\left(16 - x^{2}\right)^{-1/2} \] \[ \Rightarrow\; \frac{dy}{dx} = \frac{-x}{\sqrt{\,16 - x^{2}\,}} \] \[ \Rightarrow\; \frac{dy}{dx} = \frac{-x}{y} \]

Example

Differentiate the equation

\[ 2x^2 + 2xy + 2y^2 = 16 \]

Differentiate each term:

\[ 4x + 2\left(x\frac{dy}{dx} + y\right) + 4y\frac{dy}{dx} = 0 \]

Collect $\dfrac{dy}{dx}$ terms:

\[ (2x + 4y)\frac{dy}{dx} = -4x - 2y \]

Solve:

\[ \frac{dy}{dx} = \frac{-4x - 2y}{2x + 4y}. \]

\[ \frac{dy}{dx} = \frac{-(2x + y)}{\,x + 2y\,} \]

Gradients and Concavity

Example

Find the gradient of the tangent at the point $R(1,2)$ on the curve

\[ x^3 + y^2 = 5. \]

and determine whether the curve is concave up or concave down at this point.

Differentiating implicitly:

\[ 3x^2 + 2y\frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{3x^2}{2y}. \]

Find second derivative

\[ 3x^{2} + 2y\,\frac{dy}{dx} = 0 \] \[ 6x + 2\frac{dy}{dx}\cdot\frac{dy}{dx} + 2y\,\frac{d^{2}y}{dx^{2}} = 0 \] \[ 6x + 2\left(\frac{-3x^{2}}{2y}\right) \left(\frac{-3x^{2}}{2y}\right) + 2y\,\frac{d^{2}y}{dx^{2}} = 0 \] \[ 6x + 2\left(\frac{9x^{4}}{4y^{2}}\right) + 2y\,\frac{d^{2}y}{dx^{2}} = 0 \] \[ 6x + \frac{9x^{4}}{2y^{2}} + 2y\,\frac{d^{2}y}{dx^{2}} = 0 \] \[ 2y\,\frac{d^{2}y}{dx^{2}} = -6x - \frac{9x^{4}}{2y^{2}} \] \[ y\,\frac{d^{2}y}{dx^{2}} = -3x - \frac{9x^{4}}{4y^{2}} \]

Divide through by y

\[ \frac{d^{2}y}{dx^{2}} = -\frac{3x}{y} \;-\; \frac{9x^{4}}{4y^{3}} \] \[ \frac{d^{2}y}{dx^{2}} = \frac{-12xy^{3} - 9x^{4}y}{4y^{4}} \] \[ \frac{d^{2}y}{dx^{2}} = \frac{-3xy\left(4y^{2} + 3x^{3}\right)}{4y^{4}} \] \[ \frac{d^{2}y}{dx^{2}} = \frac{-3x\left(4y^{2} + 3x^{3}\right)}{4y^{3}} \]

Now substitute $x=1$, $y=2$ to find the particular solution

\[ \frac{dy}{dx} = -\frac{3(1)^2}{2(2)} = -\frac{3}{4}. \] \[ \text{The gradient of the curve at this point is } \frac{-3}{4} \]

To determine concavity, differentiate again (using product and chain rules):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{3x^2}{2y}\right). \]

\[ \frac{d^{2}y}{dx^{2}} = \frac{-3\left(4\times 2^{2} + 3\right)}{4\times 2^{3}} = \frac{-3\times 19}{32} \] \[ \therefore\; \text{the curve is concave down} \] \[ \text{since } \frac{d^{2}y}{dx^{2}} \lt 0 \]

Worked solution showing second derivative test for concavity on the curve x^3 + y^2 = 5