DIFFERENTIATION OF COMPOSITE FUNCTIONS

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A composite function, as I told you before, is a function that is written inside another function.

For example, if f(x)=x+1, g(x)=x².

Then the composite function f(g(x)) is

$$f(g(x))=(x)^2+1$$

To differentiate composite functions, we use the chain rule which is exemplified as

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

That is, the derivative of a composite function is simply the derivative of the outer function multiplied by the derivative of the inner function.

Example 1

Differentiate $y=\ln(\ln(x))$

Solution:

Let $u=\ln(x)$ so that $y=\ln(u)$

$\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dx}=\frac{1}{x}$

Recalled that $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$. Accordingly

$$\frac{dy}{dx}=\frac{1}{u}\times\frac{1}{x}$$

$$\frac{dy}{dx}=\frac{1}{(u)(x)}$$

Because $u=\ln(x)$

$$\frac{dy}{dx}=\frac{1}{x \ln(x)}$$

Example 2

Differentiate the function $y=\ln(\cos x)$

Solution:

Let $u=\cos x$ so that $y=\ln(u)$

$\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dx}=-\sin x$

Recalled that $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$. Accordingly,

$$\frac{dy}{dx}=\frac{1}{u}\times -\sin x$$

$$\frac{dy}{dx}=\frac{-\sin x}{u}$$

Because $u=\cos x$

$$\frac{dy}{dx}=\frac{-\sin x}{\cos x}$$

Example 3

Find the derivative of $y=\cot 27x$

Solution:

Let $u=27x$ so that $y=\cot u$

$\frac{dy}{du}=-\csc^2 u$ and $\frac{du}{dx}=27$

$$\frac{dy}{dx}=-\csc^2 u ×27$$

$$\frac{dy}{dx}=-27\csc^2u$$

Recalled that u=27x

$$\frac{dy}{dx}=-27\csc^2(27x)$$

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Example 4

Differentiate $y=\sin(\cos 3x)$

Solution:

Let $u=\cos 3x$ so that $y=\sin u$

$\frac{dy}{du}=\cos u$ and $\frac{du}{dx}=-3\sin 3x$

Note, we obtain $\frac{du}{dx}=-3\sin 3x$ using chain rule because cos 3x is a composite function of f(x)=cos x and g(x)=3x

Recall that $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$. Accordingly, 

$$\frac{dy}{dx}=\cos u\times -3\sin 3x$$

$$\frac{dy}{dx}=-3\sin(3x) \cos u$$

Recall, $u=\cos 3x$

$$\frac{dy}{dx}=-3\sin(3x) \cos(\cos 3x)$$


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