A Step into The World of Calculus: Rules of Differentiation

We can use limits to find the general formula to calculate the derivative of a function. Over time, mathematicians found the general formulas for many functions and their combinations, so it be easier to solve complex derivatives. These formulas presented as “rules of differentiation”.

Rules of Differentiation

Constant rule:

derivative of a constant is \(zero\). Like the derivative of \( 0, 2, \sqrt{5} , e, … \)

Please note that this doesn’t apply to situations where a constant is multiplied by the variables. Same as where the derivative of \( f(x)=x \) (or \( f(x)=1x \) with respect to \( x \) , is \( 1 \) , the derivative of a function like \(f(x)=3x\), would become \(3\). This is called Multiplication by constant rule

If x had an exponent, we need to use power rule.

Power rule:

if \(n\) is a constant (positive or negative), derivative of \( x^n \) is:\(nx^{n-1} \)

Meaning the value of exponent gets multiplied by the expression, and the new exponent of the variable is equal to: previous exponent, minus one

If we also had a constant before \( x^n \) , for example \( mx^n\) , we use multiplication by constant rule, and find the derivative using: \( (n \times m) \times x^{n-1} \)

For example:

\( (3x^2)’ = 6x \)

\( (-4x^3)’ = -12x^2 \)

Sum/difference rule:

derivative of sum/difference of two functions, is equal to sum/difference of their derivatives. Meaning we can operate like we normally do, when finding the derivative of \( f(x)±g(x) \)

For example, \( f(x)=2x \) and \( g(x)=5x \), therefore \(f’ (x)=2 \) and \( g’ (x)=5 \). So \( f’ (x)+g’ (x)=7 \) , or \( f(x)+g(x)=2x+7x \), which has a derivative equal to \( 7 \).

But this is not the case when facing multiplication and division derivatives.

Product rule:

to find the derivative of multiplication of two functions (for example \( f \) and \( g \) ), we need to find the derivative of the \( f \), multiply it by the \( g \), then add the result to derivative of \( g \)  multiplied by \( f \).

Or: \( (f(x) \times g(x))’ = f'(x) \cdot g(x) + g'(x) \cdot f(x) \)

Assume \( f(x) = 3x \) and \( g(x) = 4x \), so \( f(x) \times g(x) = 12x^2 \) , which has a derivative of \( 24x \).

Now we try the formula above: \( (f(x) \times g(x))’ = f'(x) \cdot g(x) + g'(x) \cdot f(x) \)

So: \( (3x \times 4x)’ = [(3x)’.(4x)] + [(4x)’.(3x)] = [3 \times 4x] + [4 \times 3x] = 12x + 12x = 24x \)

(while  \( f'(x) = 3 \) and  \( g'(x) = 4 \), and \( 3×4 \) would be \( 12\), not \(24x\) )

Quotient rule:

same as product rule, we cant find the derivative of division of two functions, by dividing their derivatives. Instead, we use this formula:

\( \left(\frac{f(x)}{g(x)}\right)’ = \frac{f'(x) \cdot g(x) – g'(x) \cdot f(x)}{[g(x)]^2} \)

If you look closely at the numerator, you can see it has a structure that’s really similar to product rule

And for the denominator, we multiply the \( g(x) \) by itself.

For example: \( f(x) = 6x^2 \), \( g(x) = 3x \), so \( f(x) / g(x) = 2x \), which has a derivative of \( 2 \) .

Or using the formula, we have:

\( \frac{(6x^2)'(3x) – (6x^2)(3x)’}{(3x)^2} = \frac{12x(3x) – (6x^2)3}{9x^2} = \frac{36x^2 – 18x^2}{9x^2} = 2 \)

Chain rule:

we saw that multiplication and division of functions can affect derivative calculations, causing them to behave different than expected. Same goes for composite of functions (for instance: \( g(f(x)) \) ).

As an example, take a look at this function: \( (3x)^2 \), which is the function \( 3x \), composited in \( x^2 \) function. To solve this kind of problem, we can’t just use the power rule and get \( 6x \) as result, one way is to simplify the expression and take the derivative of the result:

\( (3x)^2 = 9x^2 \), which has a derivative of \( 18x \).

For simple composites like this, we can easily do this, but composite functions can get quite complicated. 

That’s why we use this formula: \( (f(g(x)))’ = f'(g(x)) \cdot g'(x) \)

Meaning after taking the derivative of \( f(g(x)) \) as normal, we need to multiply the derivative of the inner function \( (g(x)) \) to the result. So for the example above:

\( ((3x)^2)’ = 6x \times (3x)’ = 6x \times 3 = 18x \)

This formula is used for one composite. Chain rule can be applied to much more complex composite functions, having several functions nested inside one another.

Here is another example: \( ((6x^2)^3)’ = 3 \times (6x^2)^2 \times 12x = 1296x^5 \)

Let’s complicate things a notch: find  \( \left(\frac{(3x)^{-2}}{(-2x)^3}\right)’ \)

Using chain rule and quotient rule, we have:

  \( \frac{[-2 \cdot (3x)^{-3} \cdot 3 \cdot (-2x)^3] – [(3x)^{-2} \cdot 3(-2x)^2 \cdot (-2)]}{[(-2x)^3]^2} = \frac{[-2 \cdot \frac{1}{27x^3} \cdot 3 \cdot (-8x^3)] – [\frac{1}{9x^2} \cdot 12x^2 \cdot (-2)]}{64x^6} = \frac{\frac{48x^3}{27x^3} – \frac{-24x^2}{9x^2}}{64x^6} = \frac{5}{72} x^{-6} \)