# What is the difference between implicit and explicit differentiation?

Sep 20, 2016

It is a difference in how the function is presented before differentiating (or how the functions are presented).

#### Explanation:

$y = - \frac{3}{5} x + \frac{7}{5}$ gives $y$ explicitly as a function of $x$.

$3 x + 5 y = 7$ gives exactly the same relationship between $x$ and $y$, but the function is implicit (hidden) in the equation. To make the function explicit, we solve for $x$

In ${x}^{2} + {y}^{2} = 25$, $y$ is not a function of $x$. However, there are two functions implicit in the equation. We can make the functions explicit by solving for $y$.

$y = \pm \sqrt{25 - {x}^{2}}$ is equivalent to the equation above and it has 2 functions that are not too difficult to make explicit:

$y = \sqrt{25 - {x}^{2}}$ gives $y$ as a function of $x$ and

$y = - \sqrt{25 - {x}^{2}}$ gives $y$ as a different function of $x$.

We can differentiate either the implicit or explicit presentations.

Differentiating implicitly (leaving the functions implicit) we get

$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ $\text{ }$ so $\text{ }$ $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

The $y$ in the formula for the derivative is the price we pay for not making the function explicit. It replaces the explicit form of the function, whatever that may be.

For $y = \sqrt{25 - {x}^{2}}$, we get $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{\sqrt{25 - {x}^{2}}}$ (use the power and chain rule), and

for $y = - \sqrt{25 - {x}^{2}}$, we get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{\sqrt{25 - {x}^{2}}}$.

The equation ${y}^{5} + 4 {x}^{2} {y}^{2} - 3 y + 7 x = 28$ cannot be solved algebraically for $y$, (or anyway, some 5th degree equations cannot be solved) but there are several functions of $x$ implicit in the equation. You can see them in the graph of the equation (shown below).

graph{y^5+4x^2y^2-3y+7x=28 [-7.14, 6.91, -4.66, 2.36]}

We can cut the graph into pieces, each of which is the graph of some function of $x$ on some domain.

Implicit differentiation allow us to find the derivative(s) of $y$ with respect to $x$ without making the function(s) explicit. Doing that, we can find the slope of the line tangent to the graph at the point $\left(1 , 2\right)$.