Sunshine Maths

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/du x du/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/du x du/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:

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
y = (x⁴ – 2x³ + 5x + 2)²

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

= (2x⁴ – 4x³ + 10x + 4).(4x³ – 6x² + 5)
y = $(3x~-~1)^{1/2}$

y’ = ${1/2}.(3x~-~1)^{{1/2}-1}.3$

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

=
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

= 4.(2x + 3)³.2 = 8(2x + 3)³

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