Differentiation from First Principles

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:

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:

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

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):

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

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.

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.

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.	\(f'(x) = \lim_{h \to 0} \frac{e^x(e^h-1)}{h} = e^x(1) = e^x\)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.