# Product Rule

## Key Questions

$y ' = \left(\frac{1}{x} ^ 2 - \frac{3}{x} ^ 4\right) \cdot \left(1 + 15 {x}^{2}\right) + \left(- \frac{2}{x} ^ 3 + \frac{12}{x} ^ 5\right) \left(x + 5 {x}^{3}\right)$

Solution :
Suppose we have $y = f \left(x\right) \cdot g \left(x\right)$
Then, using Product Rule, $y ' = f \left(x\right) \cdot g ' \left(x\right) + f ' \left(x\right) \cdot g \left(x\right)$

In simple language, keep first term as it is and differentiate the second term, then differentiate the first term and keep the second term as it is or vice-versa.

So, here if we consider,

$f \left(x\right) = \left(\frac{1}{x} ^ 2 - \frac{3}{x} ^ 4\right)$
$g \left(x\right) = \left(x + 5 {x}^{3}\right)$

Then,

$f ' \left(x\right) = \left(- \frac{2}{x} ^ 3 + \frac{12}{x} ^ 5\right)$
$g ' \left(x\right) = \left(1 + 15 {x}^{2}\right)$

Hence, using the product rule,

$y ' = \left(\frac{1}{x} ^ 2 - \frac{3}{x} ^ 4\right) \left(1 + 15 {x}^{2}\right) + \left(- \frac{2}{x} ^ 3 + \frac{12}{x} ^ 5\right) \left(x + 5 {x}^{3}\right)$

In case , if we have more than two function, let see

$y = u \left(x\right) \cdot v \left(x\right) \cdot w \left(x\right)$

then,

$y ' = u ' \left(x\right) \cdot v \left(x\right) \cdot w \left(x\right) + u \left(x\right) \cdot v ' \left(x\right) \cdot w \left(x\right) + u \left(x\right) \cdot v \left(x\right) \cdot w ' \left(x\right)$

i.e. differentiate one function at a time and keep the remaining two as it is or consider them as constant and similarly follow for the remaining two.

• The product rule states that $\left(h f g\right) ' = h ' f g + h f ' g + h f g '$.

So, $\left(\frac{d}{\mathrm{dx}} \left(5 - x\right)\right) \left(x - 3\right) \left(2 - 3 x\right) + \left(5 - x\right) \left(\frac{d}{\mathrm{dx}} \left(x - 3\right)\right) \left(2 - 3 x\right) + \left(5 - x\right) \left(x - 3\right) \left(\frac{d}{\mathrm{dx}} \left(2 - 3 x\right)\right)$

The derivative of $5 - x$ is $- 1$ since the constant has a derivative of o and the derivative of $- x$ is $- 1$ from $\left(1\right) \cdot - 1 {x}^{1 - 1}$ giving $- 1 {x}^{0} \mathmr{and} - 1.$

The derivative of $x - 3$ is 1 as above.

The derivative of $2 - 3 x$ is -3 from $1 \cdot - 3 {x}^{1 - 1} = - {3}^{0}$

Substituting back we get $= \left(- 1\right) \left(\left(x - 3\right)\right) \left(\left(2 - 3 x\right)\right) + \left(\left(5 - x\right)\right) \left(1\right) \left(\left(2 - 3 x\right)\right) + \left(\left(5 - x\right)\right) \left(\left(x - 3\right)\right) \left(- 3\right)$

Using FOIL we get $3 {x}^{2} - 11 x + 6 + 3 {x}^{2} - 17 x + 10 + 3 {x}^{2} - 24 x + 45.$

Collecting like terms we get our derivative; $9 {x}^{2} - 52 x + 61.$

• The product rule for derivatives states that given a function $f \left(x\right) = g \left(x\right) h \left(x\right)$, the derivative of the function is $f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

The product rule is used primarily when the function for which one desires the derivative is blatantly the product of two functions, or when the function would be more easily differentiated if looked at as the product of two functions. For example, when looking at the function $f \left(x\right) = {\tan}^{2} \left(x\right)$, it is easier to express the function as a product, in this case namely $f \left(x\right) = \tan \left(x\right) \tan \left(x\right)$.

In this case, expressing the function as a product is easier because the basic derivatives for the six primary trig functions ($\sin \left(x\right) , \cos \left(x\right) , \tan \left(x\right) , \csc \left(x\right) , \sec \left(x\right) , \cot \left(x\right)$) are known, and are, respectively, $\cos \left(x\right) , - \sin \left(x\right) , {\sec}^{2} \left(x\right) , - \csc \left(x\right) \cot \left(x\right) , \sec \left(x\right) \tan \left(x\right) , - {\csc}^{2} \left(x\right)$

However, the derivative for $f \left(x\right) = {\tan}^{2} \left(x\right)$ is not one of the elementary 6 trigonometric derivatives. Thus, we consider $f \left(x\right) = {\tan}^{2} \left(x\right) = \tan \left(x\right) \tan \left(x\right)$ so that we can deal with $\tan \left(x\right)$, for which we know the derivative. Utilizing the derivative of $\tan \left(x\right)$, namely $\frac{d}{\mathrm{dx}} \tan \left(x\right) = {\sec}^{2} \left(x\right)$, and the Chain Rule $\frac{\mathrm{df}}{\mathrm{dx}} = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$, we obtain:

$f ' \left(x\right) = \left[\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right)\right] \tan \left(x\right) + \tan \left(x\right) \left[\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right)\right]$

$\frac{d}{\mathrm{dx}} \tan \left(x\right) = {\sec}^{2} \left(x\right)$, so...

$f ' \left(x\right) = {\sec}^{2} \left(x\right) \tan \left(x\right) + \tan \left(x\right) {\sec}^{2} \left(x\right) = 2 \tan \left(x\right) {\sec}^{2} \left(x\right)$