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In our previous discussion of differentiation of trigonometric functions, We noted the following derivatives.

$$\frac{d}{dx}(\sin x)=\cos x$$

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

$$\frac{d}{dx}(\tan x)=\sec^2 x$$

$$\frac{d}{dx}(\sec x)=\sec x \tan x$$

$$\frac{d}{dx}(\csc x)=-\csc x \cot x$$

$$\frac{d}{dx}(\cot x)=-\csc^2 x$$

But, How can we find the derivative of composite function such as $\cos 3x$, $\sec 4x$, $\sin (3x)$? We use the chain rule which states that:


To put it in words, the chain rule of differentiation states that when differentiating a composite function, you must first differentiate the outer function before differentiating the inner function.

To better appreciate this rule, let take some examples.

Example 1

Differentiate $\cos 3x$


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

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

Example 2

Differentiate $y=12\cos 3x$


$$\frac{dy}{dx}=12(-\sin 3x) \times 3$$

$$\frac{dy}{dx}=-36\sin 3x$$

Example 3

Differentiate the function $h(x)=4\sin 8x-12\cos 3x-\cot 3x$


$h'(x)=4\cos 8x\times 8-(12 (-\sin 3x)\times 3)-(-csc^23x\times 3)$

$h'(x)=32\cos 8x+36\sin 3x+3\csc²(3x)$

Example 4

Differentiate $y=x^3\sin 3x$


$y=x^3\sin 3x$ is the product of x³ and $\sin 3x$. Accordingly, we will use the product rule exemplified as:


Here, $f=x^3$, $f'=3x^2$, $g=\sin 3x$, $g'=3\cos 3x$

$$\frac{dy}{dx}=3x^2(\sin 3x)+(3 \cos 3x)(x^3)$$

$$\frac{dy}{dx}=3x^2\sin 3x+3x^3 \cos 3x$$

$$\frac{dy}{dx}=3x^2(\sin 3x+x\cos 3x)$$

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Example 5

If $y=x^2\cot 2x$, Find $\frac{dy}{dx}$


Again, $x^2\cot 2x$ is a product, therefore, we use the product rule.

Here, $f=x^2$, $f'=2x$, $g=\cot 2x$, $g'=-2\csc^2 x$

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

$$\frac{dy}{dx}=2x \cot 2x-2x^2 \csc^2 x$$

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

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