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Differentiation is the process of finding the gradient of a variable function. A variable function is a polynomial function that takes the shape of a curve, so is therefore a function that has an always-changing gradient.
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Team Differentiation from First Principles Teachers
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.
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 below is often found in the formula booklets that are given to students to learn differentiation from first principles:
\[f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\]
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:
\[\frac{d}{dx} \sin x = \cos x\]
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:
\[\frac{d}{dx} (\cos x) = -\sin x\]
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 x and x + h. | 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 y = f(x) be a function. Pick two points x and x + h. | 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. | |
| ANSWER | \(e^x\), but of course, the entire proof is an answer as this is differentiation from first principles. |
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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.
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