# Derivatives of functions

Maybe you learned about functions a while ago. Maybe you’ve even come to know and love them! If you have, then you know that some simpler functions (like a linear function) can be easily examined and graphed. But what do we do when we need to graph more complicated functions?

Enter: First derivatives!

Since the derivation is a rate of change of a function, you can determine if the function is increasing or decreasing. (BTW: If the rate of change of the function is 0, that’s where the function does not increase or decrease, so then the function has critical points.)

Ready to get into it?

## What does it mean to find the derivative of a function?

A derivative of a function is a rate of change of the function with respect to change in a variable. Actually, to find the first derivative of the function $$f(x_0)$$ means to determine the slope of the tangent line to the graph of the function at $$x_0$$.

To simplify the process of differentiation, we use differentiation rules rather than the definition of the derivative. There are quite a few, so take a look and keep them handy as we get started:

Constant multiple property of derivatives | $$\frac{d}{dx}\left(c\times f(x)\right)=c\times\frac{d}{dx}\left(f(x) \right)$$ |

Sum rule for derivatives | $$\frac{d}{dx}\left(f(x) + g(x)\right)=\frac{d}{dx}\left( f(x) \right)+\frac{d}{dx}\left( g(x) \right)$$ |

Difference rule for derivatives | $$\frac{d}{dx}\left(f(x) - g(x)\right)=\frac{d}{dx}\left( f(x) \right)-\frac{d}{dx}\left( g(x) \right)$$ |

Product rule for derivatives | $$\frac{d}{dx}\left(f(x)\times g(x)\right)=\frac{d}{dx}\left( f(x) \right)\times g(x)+f(x)\times\frac{d}{dx}\left( g(x) \right) $$ |

Quotient rule for derivatives | $$\frac{d}{dx}\left(\frac{f(x)}{g(x)} \right)=\frac{\frac{d}{dx}\left(f(x) \right)\times g(x)-f(x)\times\frac{d}{dx}\left( g(x) \right)}{(g(x))^{2}}$$ |

The Chain rule | $$(f\circ g)^{\prime}(x)=f^{\prime}(g(x))\times g^{\prime}(x) \text{ or }\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\text{ when }y=f(u), u=g(x)$$ |

Derivative of the inverse function | $$\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}\left(x\right)\right)}$$ |

### Why is the first derivative so useful?

You may already know that functions can be really important, as they can represent lots of real-life situations. Derivatives are closely connected to the functions and also have many useful applications!

Derivatives represent the rate of change, so that means the velocity, acceleration, and some other physical entities can be calculated using derivatives. By comparing the slope of the tangents to the function at a certain interval, you can see if the function is increasing or decreasing on that interval. If you find the second derivative of a function, you can determine if the function is concave (up or down) on the interval.

## How to find the derivative

Let’s dive right into some examples, which we’ll walk through together!

### Example 1

**Find the derivative of the function:**

Take the derivative of both sides:

$$f^{\prime}(x)={\frac{d}{dx}}(2x^{2} + 3x – 4)$$

Use differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:

$$f^{\prime}(x)={\frac{d}{dx}}({2}x^{2}) +\frac{d}{dx} ({3}x) – \frac{d}{dx}(4)$$

Use differentiation rule $$\frac{d}{dx}({a}\times f)={a}\frac{d}{dx}(f)$$:

$$f^{\prime}(x)={{2}\frac{d}{dx}}(x^{2}) +{3}\frac{d}{dx} (x) – \frac{d}{dx}(4)$$

Next, use $$\frac{d}{dx}(x^{n})={n}\times x^{{n}-1}$$:

$$f^{\prime}(x)={2}\times {2}x^1 +{3}{\frac{d}{dx} (x) }- \frac{d}{dx}(4)$$

Use $$\frac{d}{dx} x=1$$:

$$f^{\prime}(x)={2}\times {2}{x^1} +{3\times}{1 }- \frac{d}{dx}(4)$$

The derivative of any constant is equal to zero, since the derivative represents the rate of change and the constant function does not change. That means the derivative of $$4$$ is equal to $$0$$:

$$f^{\prime}(x)={2}\times {2}{x^1} +{3\times}{1 }- 0$$

Simplify the expressions:

$$f^{\prime}(x)={{2}\times {2}{x}} +3- 0$$

Simplify the term and remove the zero (remember: subtracting zero doesn’t change the value):

$$f^{\prime}(x)={4x} +3$$

We found the derivative of a function $$f(x)=2x^{2} + 3x – 4$$ :

Woohoo! Let’s do another one just to make sure.

### Example 2

**Find the derivative of the function:**

Take the derivative of both sides:

$$f^{\prime}(x)={\frac{d}{dx}}\left(\frac{x}{\cos{x}}\right)$$

Use differentiation rule $$\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{\frac{d}{dx}\left({f}\right)\times {g}-{f}\times \frac{d}{dx}\left({g}\right)}{{g}^2}$$:

$$f^{\prime}(x)=\frac{\frac{d}{dx}({x})\times {\cos{x}}-{x}\times \frac{d}{dx}({\cos{x}})}{({\cos{x})}^2}$$

Use $$\frac{d}{dx} x=1$$:

$$f^{\prime}(x)=\frac{{1}\times {\cos{x}}-{x}\times {\frac{d}{dx}(\cos{x})}}{(\cos{x})^2}$$

Next, use $$\frac{d}{dx}(\cos{x})=-\sin{x}$$:

$$f^{\prime}(x)=\frac{{{1}\times {\cos{x}}}-{x}\times {(-\sin{x})}}{(\cos{x})^2}$$

Simplify the expression on the right side:

$$f^{\prime}(x)=\frac{ {\cos{x}}+{{x}\times {\sin{x}}}}{(\cos{x})^2}$$

We’ve found the derivative of a function $$f(x)=\frac{x}{\cos{x}}$$:

That wasn’t so bad, right? Now that we’ve walked through some detailed examples, let’s review the overall process so you can learn how to use it with *any* problem:

## Study summary

- Take the derivative on both sides of the equation.
- Use the differentiation rules.
- Find the derivative.
- Simplify the expression, if possible.

## Do it yourself!

How did you do when we walked through the examples together? If you think you need more practice, try a few of these!

**Take the derivative of a function:**

- $$f(x)=10x^4+x$$
- $$f(x)=e^x+5x$$
- $$f(x)=\log_5x$$
- $$f(x)=\frac{x-1}{\sqrt{4x^2-1}}$$

*Solutions:*

- $$f^{\prime}(x)=40x^3+1$$
- $$f^{\prime}(x)=e^x+5$$
- $$f^{\prime}(x)=\frac{1}{x\ln{5}}$$
- $$f^{\prime}(x)=\frac{-1+4x}{\sqrt{4x^2-1}(4x^2-1)}$$

If you’re struggling through the solving process, that’s normal! Scan the problem using your Photomath app, and we’ll walk you through to the other side!

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