## The Chain Rule

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# The Chain Rule

The chain rule is one of the most important rules for finding derivatives. Before we state it, let us suppose that y is a function of u:

y = f (u)

and u in turn is a function of x:

u = g (x)

Then by substituting g (x) for u in the first function, we can consider y as a function of x:

y = f (g (x)).

Functions of this type are called composite functions. To find the derivative of f (g(x)), we could, of course, explicitly set up f (g (x)) and then differentiate it. Very often, the algebra we would have to go through to find f (g (x)) explicitly is very cumbersome and we want to avoid it. Fortunately, the chain rule will allow us to differentiable f (g (x)) without explicitly knowing f (g (x)).

dv/dx = dv/du . du/dx.

By the definition of the derivative,

dv/dx = lim Δy/Δx

Δx→0

= lim [Δy/Δu . Δu/Δx]

Δx→0

= lim Δy/Δu . lim Δu/Δx

Δx→0

Now, u, being differentiable function of x, is continuous. Hence Δu→0 as Δx→0. Thus

Δy/Δx = lim Δy/Δu . lim Δu/ Δx

Δx→0 Δx→0

= dy/du . du/dx .

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y = f (u)

and u in turn is a function of x:

u = g (x)

Then by substituting g (x) for u in the first function, we can consider y as a function of x:

y = f (g (x)).

Functions of this type are called composite functions. To find the derivative of f (g(x)), we could, of course, explicitly set up f (g (x)) and then differentiate it. Very often, the algebra we would have to go through to find f (g (x)) explicitly is very cumbersome and we want to avoid it. Fortunately, the chain rule will allow us to differentiable f (g (x)) without explicitly knowing f (g (x)).

**Rule (chain Rule)**. If y = f (u) is a differentiable function of u and u = g (x) is a differentiable function of x, then y = f (g (x)) is a differentiable function of x anddv/dx = dv/du . du/dx.

**Proof.**Let Δu denote a change in u as determined from u = g (x) corresponding to a change Δu in u, let Δy be the change in y as determined from y = f (u).By the definition of the derivative,

dv/dx = lim Δy/Δx

Δx→0

= lim [Δy/Δu . Δu/Δx]

Δx→0

= lim Δy/Δu . lim Δu/Δx

Δx→0

Now, u, being differentiable function of x, is continuous. Hence Δu→0 as Δx→0. Thus

Δy/Δx = lim Δy/Δu . lim Δu/ Δx

Δx→0 Δx→0

= dy/du . du/dx .

For more help in

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