#worked example

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Solve With Inverse Matrix -- Example

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Calculate Human Power -- Example

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Implicit Differentiation

Implicit differentiation of equations

Cont’d from “Worked examples: Example 2”

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A process is followed to implicitly differentiate a function. We’ll start by considering the equation:

a + f(y(x)) = g(x)

Since we have a function of y it is difficult (or impossible depending on the function) to rearrange to be solely in terms of x on the right hand side.

Now, let’s recall what we learnt about the differential operator (See “The differential operator”). To first differentiate any equation we apply the differential operator to both sides of the equation. So, if we do this the equation becomes

d/dx [a + f(y(x))] = d/dx [g(x)]

which represents the first derivative of our original equation. Now, using the summation rule of differentiation, which states that d/dx (f(x) + g(x)) = d/dx f(x) + d/dx g(x), the left hand side (LHS) of the equation becomes

d/dx [a + f(y(x))] = d/dx a + d/dx f(y(x))

and since we know that the derivative of any constant (i.e. something that is independent of x) is zero; d/dx a = 0, Therefore this equation becomes

d/dx f(y(x)) = d/dx g(x).

We can now apply the chain rule to the LHS in the same way we did in example 2, to get

d/dx f(y(x)) = d/dy f(y(x)) d/dx y(x)

which we can substitute back in to the equation to get

dy/dx d/dy f(y) = d/dx g(x)

⇒ dy/dx = d/dx g(x) [d/dy f(y)]−1

Again, this is probably best shown with an explicit example (See “Worked examples: Example 3“).