Derivative Rules

In calculus, a derivative can be thought of as an instantaneous rate of change; that is, how much a quantity is changing at a given point. Let’s take a closer look at how we can differentiate a function easily by the use of some helpful rules.

Understanding the Derivative



Differentiation is a method to compute the rate at which a dependent variable y changes with respect to the change in the independent variable $$x$$. This rate of change is called the derivative of $$y$$ with respect to $$x$$. There are many different notations to denote “take the derivative of.” The relationship between $$y$$ and $$x$$ is usually denoted by $$f(x)$$ and its derivative is usually denoted as $$f'(x)$$ or $$y’$$ or $$\frac{dy}{dx}$$. The definition of the derivative is given by:

$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$

which is a lengthy procedure used to evaluate the derivative of a function. Thankfully, easier methods have been developed to help evaluate derivatives more quickly.

The Power Rule

If n is any real number, then $$\frac{d(x^n)}{dx}=nx^{n-1}$$.

Also, remember that the derivatives of a constant is zero:  $$\frac{d(c)}{dx}=0$$.

Note that the sum and difference rule states:

$$\frac{d}{dx}[f(x)+g(x)-h(x)] = \frac{d}{dx}f(x)+\frac{d}{dx}g(x)-\frac{d}{dx}h(x)$$

(Just simply apply the power rule to each term in the function separately).

Example: Find the derivative of

$$f(x)=4-4\sqrt{x}+\frac{2}{x}$$

Solution: First, rewrite the function so that all variables of $$x$$ have an exponent in the numerator:

$$f(x)=4-4x^{\frac{1}{2}} + 2x^{-1}$$

Now, apply the power rule to the function:

$$f'(x)=0-4\cdot\frac{1}{2}x^{\frac{-1}{2}}+2\cdot(-1)x^{-2}$$

$$f'(x)=-\frac{2}{\sqrt{x}} - \frac{2}{x^2}$$

The Product Rule

We already know how to find the derivative of a sum or difference of functions, but what about the product of two functions? The product of two functions is when two functions are being multiplied together. If $$f$$ and $$g$$ are both differentiable, then the product rule states:

$$\frac{d}{dx}[f(x)g(x)] = \frac{d}{dx}[f(x)]g(x) + f(x)\frac{d}{dx}[g(x)]$$

Example: Find the derivative of h(x) = (3x + 1)(8x4 +5x).

Solution: Using the above formula, let $$f(x) = (3x+1)$$ and let $$g(x) = (8x^4 + 5x)$$.

Now, find

$$\frac{d}{dx}[f(x)]$$ and $$\frac{d}{dx}[g(x)]: \frac{d}{dx}[f(x)]=3$$ and $$\frac{d}{dx}[g(x)] = 32x^3+5$$

Lastly, apply the product rule using the above formula:

$$h'(x) = \frac{d}{dx}[(3x+1)(8x^4+5x)] = \frac{d}{dx}[f(x)]g(x)+f(x)\frac{d}{dx}[g(x)]$$

$$h'(x) = (3)(8x^4+5x)+(3x+1)(32x^3+5)$$

The Quotient Rule

Now, let's take a look at the quotient of functions (when two functions are being divided by each other). If $$f$$ and $$g$$ are both differentiable, then the quotient rule states:

$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\frac{d}{dx}[f(x)]g(x)-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^2}$$

Example: Find the derivative of

$$h(x) = \frac{3x-4}{2x^2-1}$$

Solution: Using the above formula, let $$f(x) = (3x-4)$$ and let $$g(x) = (2x^2-1)$$.

Now, find

$$\frac{d}{dx}[f(x)]$$ and$$\frac{d}{dx}[g(x)]:$$  $$\frac{d}{dx}[f(x)]=3$$ and $$\frac{d}{dx}[g(x)]=4x$$

Lastly, apply the quotient rule using the above formula:

$$h'(x) = \frac{d}{dx}\left[\frac{3x-4}{2x^2-1}\right] = \frac{\frac{d}{dx}[f(x)]g(x)-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^2}$$

$$h'(x) = \frac{(3)(2x^2-1)-(3x-4)(4x)}{[(2x^2-1)]^2}$$

The Chain Rule

If $$f$$ and $$g$$ are both differentiable and $$F$$ is the composite function defined by $$F(x) = f(g(x))$$, then $$F$$ is differentiable and the chain rule can be applied to $$F'$$, which is given by $$F '(x) = f '(g(x))g '(x)$$

Example: Find the derivative of  $$F(x) = (3x^4+2x^2-7)^5$$.

Solution: Using the above formula, let  $$g(x) = (3x^4 + 2x^2 - 7)$$ and let $$f(x) = (x)^5$$.

Now, find  $$f '(x)$$ and $$g'(x)$$: $$f'(x) = 5x^4$$ and $$g'(x) = 12x^3 + 4x - 0$$

So, if   $$f'(x) = 5x^4$$, then the composite function $$f'(g(x)) = 5(3x^4 + 2x^2 - 7)^4$$

Lastly, apply the chain rule using the above formula:

$$F'(x) = f'(g(x))g'(x)= 5(3x^4+2x^2-7)^4(12x^3+4x)$$.