Lesson 5, Topic 5
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Functions of Function (Composite Function)

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Consider :- $$\scriptsize y = \normalsize \frac{1}{(3x\; – \;4)^2 }\scriptsize = (3x -4)^{-2}$$

A function such as this sets a problem for differentiation.

It cannot be expressed as a polynomial. So we look at a very important technique devised to handle such functions.

To find $$\frac{dy}{dx}$$ for the function above, we treat $$\scriptsize = (3x -4)^{-2} \! \! \!$$ as a composite function built up in two stages from a core function (3x – 4), which we call u and then taking $$\scriptsize y = u^{-2}$$ So we have u = (3x – 4)

Hence the general form of a function of function rule also called chain rule which is given as

$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$

Then in the above problem,

$$\scriptsize y = (3x -4)^{-2}$$

Solution

$$\scriptsize y = u^{-2}, u = (3x -4)$$

Thus $$\frac{dy}{du} = \scriptsize -2u^{-3} \normalsize = \frac{-2}{u^3}$$

while $$\frac{du}{dx} \scriptsize = 3$$

$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$

$$\frac{dy}{dx} = \; – \frac{2}{u^3} \times \scriptsize 3 =\; – \normalsize\frac{6}{u^3}$$

Substituting for u

= $$– \frac{6}{(3x \;-\; 4)^3}$$

Example:

2. Differentiate $$\scriptsize y = (4x -5)^6$$

Let $$\scriptsize y = u^{6}, u = (4x – 5)$$

Thus $$\frac{dy}{du} = \scriptsize 6u^{5}$$

while $$\frac{du}{dx} \scriptsize = 4$$

$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$

= $$\scriptsize 6u^{5} \times 4$$

$$\frac{dy}{dx} = \scriptsize 24u^{5}$$

and substituting u = 4x – 5 gives $$\frac{dy}{dx} = \scriptsize 24(4x-5)^{5}$$

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