# PRODUCT RULE OF DIFFERENTIATION

The product rule of differentiation states that the derivative of the product of two functions is the sum of the derivative of the first function multiply by the second function, and the derivative of the second function multiply the first function.

That is,

$$(fg)'=f'g+g'f$$

Using this fact, we solve the following examples.

Example 1

Find the derivative of $y=(x^2+1)(x^3+1)$

Solution:

$$(fg)'=f'g+g'f$$

Here, $f=x^2+1$, $f'=2x$$, g=x^3+1, g'=3x^2 Using the formula, \frac{dy}{dx}=2x(x^3+1)+3x^2(x^2+1) \frac{dy}{dx}=2x^4+2x+3x^4+3x^2 \frac{dy}{dx}=5x^4+3x^2+2x Example 2 Differentiate y=(x^4+2x)(x^3+8x^2) Solution$$(fg)'=f'g+g'f$$Here, f=x^4+2x, f'=4x^3+2, g=x^3+8x^2, g'=3x^2+16x \frac{dy}{dx}=(4x^3+2)(x^3+8x^2)+(3x^2+16x)(x^4+2x) \frac{dy}{dx}=4x^6+32x^5+2x^3+16x^2+3x^6+6x^3+16x^5+32x^2 \frac{dy}{dx}=7x^6+48x^5+8x^3+48x^2 Example 3 Differentiate t=x^2\cot 2x Solution x^2\cot 2x is the same as x^2×\cot 2x$$(fg)'=f'g+g'f$$Here, f=x^2, f'=2x, g=\cot 2x, g'=-2\csc^22x \frac{dt}{dx}=2x(\cot 2x)+(-2\csc^2 2x)(x^2) \frac{dt}{dx}=2x\cot2x-2x^2\csc^22x \frac{dt}{dx}=2x(\cot 2x-x\csc² 2x) RELATED POSTS Example 4 Differentiate y=(x-1)(x+1)^2 Solution The above expression can be rewritten as:$$(x-1)(x+1)(x+1)$$For a three function product, we solve using$$fgh=f'gh+g'fh+h'fg$$f=x-1, f'=1$$, g=x+1$, $g'=1$, $h=x+1$, $h'=1$

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

$\frac{dy}{dx}=x^2+2x+1+x^2-1+x^2-1$

$\frac{dy}{dx}=3x^2+2x-1$

With these examples, you should be able to differentiate the product of functions. Meanwhile, if you have got questions relating to this post, feel free to ask our telegram community.
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