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Chain Rule

The Chain Rule, also called the composite function, is applicable to a function that is made up of another function. For example –

y = (2x + 3)2 can be treated as a composite of two functions.

Let f(x) = 2x + 2  = u

Then y = u2

Therefore dy/dx  =   dy/dudu/dx

This way of differentiation is called the Chain Rule.

Now differentiating our function using this rule, we get

u = 2x + 3                                 y = u2

du/dx = 2                              dy/du = 2u

dy/dx  =   dy/dudu/dx

            = 2u x 2  = 4u

            = 4(2x + 3)

            = 8x + 12

In general, {d/dx}{g(f(x))}  = g’f(x)  x  f'(x),  and

{d/dx}{(f(x))^n}  =  nf'(x)(f(x))^{n-1}

 

Now let us look at a few examples. Differentiate the following functions:

  1. y = 3 (5x + 4)4

    {d/dx}{(f(x))^n}  =  nf'(x)(f(x))^{n-1}

    y’  = 3×4 (5x + 4)3 x 5  =  60(5x + 4)3

  2.  y =  (x4 – 2x3 + 5x + 2)2

    y’ = 2.(x4 – 2x3 + 5x + 2)1.(4x3 – 6x2 + 5)

       = (2x4 – 4x3 + 10x + 4).(4x3 – 6x2 + 5)

  3. y = (3x~-~1)^{1/2}

    y’ = {1/2}.(3x~-~1)^{{1/2}-1}.3

        = {3/2}(3x~-~1)^{-1/2}

        = 3/{2 sqrt {3x~-~1)}

  4. Find the equation of the tangent to y = (2x + 3)4 at the point x = -1

    When x = -1, y = (2x-1 + 3)4 = (-2 + 3)4  =  1   (-1, 1)

    y = (2x + 3)4

    dy/dx  = 4.(2x + 3)3.2    =  8(2x + 3)3

    dy/dx at x = -1  =  m = 8(2 x -1 + 3)3  =  8

    Equation of the tangent at (-1, 1) is

    (y – 1)  =  8(x + 1)

    8x – y + 8 + 1 = 0

    8x – y + 9  = 0