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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 \[\frac{dy}{dx}.\]

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

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

 

 

Read 1971 times Last modified on Saturday, 08 February 2014 16:39
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