There is a traditional method to differentiate functions, however, we will be concentrating on finding the gradient still through differentiation but from first principles. This means using standard Straight Line Graphs methods of \(\frac{\Delta y}{\Delta x}\) to find the gradient of a function.

## How does differentiation from first principles work?

Differentiation from first principles involves using \(\frac{\Delta y}{\Delta x}\) to calculate the gradient of a function. We will have a closer look to the step-by-step process below:

STEP 1: Let \(y = f(x)\) be a function. Pick two points *x* and \(x+h\).

The coordinates of x will be \((x, f(x))\) and the coordinates of \(x+h\) will be (\(x+h, f(x + h)\)).

STEP 2: Find \(\Delta y\) and \(\Delta x\).

\(\Delta y = f(x+h) - f(x); \Delta x = x+h-x = h\)STEP 3: Complete \(\frac{\Delta y}{\Delta x}\).$$\frac{\Delta y}{\Delta x} = \frac{f(x+h) - f(x)}{h}$$STEP 4: Take a limit:\[f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\].

## The formula for differentiation from first principles

The formula below is often found in the formula booklets that are given to students to learn differentiation from first principles:

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

## Derivative of sin(x) using first principles

To find out the derivative of sin(x) using first principles, we need to use the formula for first principles we saw above:

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

Here we will substitute f(x) with our function, sin(x):

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

Using the trigonometric identity, we can come up with the following formula, equivalent to the one above:

\[f'(x) = \lim_{h\to 0} \frac{(\sin x \cos h + \sin h \cos x) - \sin x}{h}\]

We can now factor out the \(\sin x\) term:

\[\begin{align} f'(x) &= \lim_{h\to 0} \frac{\sin x(\cos h -1) + \sin h\cos x}{h} \\ &= \lim_{h \to 0}(\frac{\sin x (\cos h -1)}{h} + \frac{\sin h \cos x}{h}) \\ &= \lim_{h \to 0} \frac{\sin x (\cos h - 1)}{h} + lim_{h \to 0} \frac{\sin h \cos x}{h} \\ &=(\sin x) \lim_{h \to 0} \frac{\cos h - 1}{h} + (\cos x) \lim_{h \to 0} \frac{\sin h}{h} \end{align} \]

here we need to use some standard limits: \(\lim_{h \to 0} \frac{\sin h}{h} = 1\), and \(\lim_{h \to 0} \frac{\cos h - 1}{h} = 0\).

Using these, we get to:

\[f'(x) = 0 + (\cos x) (1) = \cos x\]

And so:

\[\frac{d}{dx} \sin x = \cos x\]

## Derivative of cos(x) using first principles

To find out the derivative of cos(x) using first principles, we need to use the formula for first principles we saw above:

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

Here we will substitute f(x) with our function, cos(x):

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

For the next step, we need to remember the trigonometric identity: \(cos(a +b) = \cos a \cdot \cos b - \sin a \cdot \sin b\).

Using the trigonometric identity, we can come up with the following formula, equivalent to the one above:

\[f'(x) = \lim_{h\to 0} \frac{(\cos x\cdot \cos h - \sin x \cdot \sin h) - \cos x}{h}\]

We can now factor out the \(\cos x\) term:

\[f'(x) = \lim_{h\to 0} \frac{\cos x(\cos h - 1) - \sin x \cdot \sin h}{h} = \lim_{h\to 0} \frac{\cos x(\cos h - 1)}{h} - \frac{\sin x \cdot \sin h}{h}\].

Now we need to change factors in the equation above to simplify the limit later. For this, you'll need to recognise formulas that you can easily resolve.

The equations that will be useful here are: \(\lim_{x \to 0} \frac{\sin x}{x} = 1; and \lim_{x_to 0} \frac{\cos x - 1}{x} = 0\)

If we substitute the equations in the hint above, we get:

\[\lim_{h\to 0} \frac{\cos x(\cos h - 1)}{h} - \frac{\sin x \cdot \sin h}{h} \rightarrow \lim_{h \to 0} \cos x (\frac{\cos h -1 }{h}) - \sin x (\frac{\sin h}{h}) \rightarrow \lim_{h \to 0} \cos x(0) - \sin x (1)\]

Finally, we can get to:

\[\lim_{h \to 0} \cos x(0) - \sin x (1) = \lim_{h \to 0} (-\sin x)\]

Since there are no more *h *variables in the equation above, we can drop the \(\lim_{h \to 0}\), and with that we get the final equation of:

\[\frac{d}{dx} (\cos x) = -\sin x\]

## Worked examples of differentiation from first principles

Let's look at two examples, one easy and one a little more difficult.

Differentiate from first principles \(y = f(x) = x^3\).

SOLUTION:

Steps | Worked out example |

STEP 1: Let \(y = f(x)\) be a function. Pick two points | Coordinates are \((x, x^3)\) and \((x+h, (x+h)^3)\). We can simplify \((x+h)^3 = x^3 + 3x^2 h+3h^2x+ h^3\) |

STEP 2: Find \(\Delta y\) and \(\Delta x\). | \(\Delta y = (x+h)^3 - x = x^3 + 3x^2h + 3h^2x+h^3 - x^3 = 3x^2h + 3h^2x + h^3; \\ \Delta x = x+ h- x = h\) |

STEP 3:Complete \(\frac{\Delta y}{\Delta x}\) | \(\frac{\Delta y}{\Delta x} = \frac{3x^2h+3h^2x+h^3}{h} = 3x^2 + 3hx+h^2\) |

STEP 4: Take a limit. | \(f'(x) = \lim_{h \to 0} 3x^2 + 3h^2x + h^2 = 3x^2\) |

ANSWER | \(3x^2\) however the entire proof is a differentiation from first principles. |

So differentiation can be seen as taking a limit of a gradient between two points of a function. You will see that these final answers are the same as taking derivatives.

Let's look at another example to try and really understand the concept. This time we are using an exponential function.

Differentiate from first principles \(f(x) = e^x\).

SOLUTION:

Steps | Worked out example |

STEP 1: Let | Co-ordinates are \((x, e^x)\) and \((x+h, e^{x+h})\). |

STEP 2: Find \(\Delta y\) and \(\Delta x\) | \(\Delta y = e^{x+h} -e^x = e^xe^h-e^x = e^x(e^h-1)\)\(\Delta x = (x+h) - x= h\) |

STEP 3:Complete \(\frac{\Delta y}{\Delta x}\) | \(\frac{\Delta y}{\Delta x} = \frac{e^x(e^h-1)}{h}\) |

STEP 4: Take a limit. | \(f'(x) = \lim_{h \to 0} \frac{e^x(e^h-1)}{h} = e^x(1) = e^x\)Because \(\lim_{h \to 0} \frac{(e^h-1)}{h} = 1\) |

ANSWER | \(e^x\), but of course, the entire proof is an answer as this is differentiation from first principles. |

## Differentiation from First Principles - Key takeaways

- Differentiation is the process of finding the gradient of a curve.
- The gradient of a curve changes at all points.
- Differentiation can be treated as a limit tending to zero.
- The formula to differentiate from first principles is found in the formula booklet and is \(f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

###### Learn with 2 Differentiation from First Principles flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Differentiation from First Principles

How do we differentiate from first principles?

We take the gradient of a function using any two points on the function (normally x and x+h).

What is the differentiation from the first principles formula?

The formula is:

limh->0((f(x+h)-f(x))/h)

How do we differentiate a quadratic from first principles?

We simply use the formula and cancel out an h from the numerator. This should leave us with a linear function.

How do we differentiate a trigonometric function from first principles?

We use addition formulae to simplify the numerator of the formula and any identities to help us find out what happens to the function when h tends to 0.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more