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Tuesday, 13 November 2012 20:24

Implicit Differentiation

By  Gideon
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Implicit Differentiation

Implicit Differentiation involves finding the derivative of a function $y=f(x),$by first finding the derivative of a function $g\left(x, y(x) \right),$and then solving for $\frac{dy}{dx}.$

As such, it involves both the chain rule for differentiation, and (potentially), the rule for differentiating a product.

First, recall that the chain rule states that if $g$is a function of $y(x),$then:

$\frac{d}{dx} g\left( y(x) \right) = \frac{dg}{dy} \frac{dy}{dx}.$

So, for example, if:

$y(x) = x^{2}$

$g(y) = 3y + 2$

then:

$g\left( y(x)\right) = 3x^{2} + 2$

$\frac{dy}{dx} = 2x,$

$\frac{dg}{dy} = 3,$

so that:

$\frac{dg}{dx} = \frac{dg}{dy} \frac{dy}{dx} = 3(2x) = 6x.$

The product rule, on the other hand, states that if $f(x)$, and $g(x)$are two functions of $x$, then:

$\frac{d}{dx}\left(f(x)g(x) \right) = \left( \frac{df}{dx} \right) g(x) + f(x) \left( \frac{dg}{dx} \right).$

For example, if:
$f(x) = x^{3} + 1$

$g(x) = 2x$

then:

$\frac{df}{dx} = 3x^{2}$

$\frac{dg}{dx} = 2$

so that:

$\frac{d}{dx} \left( f(x)g(x) \right) = 3x^{2}(2x) + (x^{3} + 1)(2)$

$= 6x^{3} + 2x^{3} + 2 = 8x^{3} + 2,$

which we can check by differentiating:

$\frac{d}{dx}\left( f(x)g(x) \right) = \frac{d}{dx}(2x^{4} + 2x) = 8x^{3} + 2.$

Now that we have these preliminaries out of the way, we are ready to tackle implicit differentiation.

Suppose we are given, for example, the equation:

$x^{2} + xy^{2} = -1.$

First we put this in the form $g(x, y(x)) = 0.$

In other words:

$x^{2} + xy^{2} + 1= 0.$

Now we differentiate both side with respect to $x.$

We obtain:
$\frac{d}{dx}x^{2} = 2x,$

$\frac{d}{dx}\left( xy^{2} \right) = y^{2} + x2y\frac{dy}{dx},$(by the product rule and the chain rule),

$\frac{d}{dx}(1) = 0,$

$\frac{d}{dx}(0) = 0,$

so that:

$\frac{d}{dx} ( x^{2} + xy^{2} + 1) = 2x + y^{2} + x2y\frac{dy}{dx} = 0.$

Now we solve this equation for $\frac{dy}{dx}.$

$2xy\frac{dy}{dx} = -2x -y^{2},$

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