# A Step into The World of Calculus: Rules of Differentiation

Differentiation rules are fundamental tools in calculus, providing efficient methods to determine the derivative of a function. These rules, including the power, product, quotient, and chain rules, simplify the process of finding rates of change and slopes of curves. Mastering these rules is essential for solving a wide range of mathematical problems.

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:

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

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