Continuity and Differentiability

Algebra of Derivatives

After learning how to define the derivatives of functions and its significance, now it’s time to move on to applying the derivatives in simple formulae and equations. The branch of the algebra of derivatives deals with exactly this task at hand. We’ll discuss certain rules on how to take derivatives of different quantities, based on the type of equations (addition or multiplication etc.) they occur in, and see some solved examples on the same. Let’s dive straight into it!

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Algebra of Derivaties

Let us have two differentiable functions f(x) and g(x) with a common domain.In the theorems that will follow, we’ll discuss how to take the derivatives of these functions when they occur in different types of equations. To prove the theorems, we’ll need to keep the definition of the derivative of a function in mind i.e. $$ {\frac{df}{dx} = Lim_{h\rightarrow{0}}\frac{f(x + h) – f(x)}{h}} $$

Derivatives

Theorem 1: The derivative of the sum of two functions is the sum of the derivatives of the functions.

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

Proof: Using the definition of the derivatives, one can see that:
$$ {\frac{d}{dx}f(x) + \frac{d}{dx}g(x) = Lim_{h\rightarrow{0}}\frac{f(x + h) – f(x)}{h} + Lim_{h\rightarrow{0}}\frac{g(x + h) – g(x)}{h}} $$

Using the summation rule of limits on the right-hand side of the equation, we get:
$${ R.H.S. = Lim_{h\rightarrow{0}}\frac{f(x + h) – f(x) + g(x + h) – g(x)}{h}} $$
$${ Lim_{h\rightarrow{0}}\frac{[f(x + h) + g(x + h)] – [f(x) + g(x)]}{h}} $$
$${ \frac{d}{dx}[f(x) + g(x)] }$$
Hence Proved.

Theorem 2: The derivative of the difference of two functions is the difference of the derivatives of the functions.

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

The proof of this theorem can be stated in a similar fashion to the proof of the first theorem.

Theorem 3: The derivative of the product of two functions is given by the Product Rule.

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

Proof: Let us begin from the definition of the derivative of the function f(x).g(x).
$${ \frac{d}{dx}(f(x)g(x)) = Lim_{h\rightarrow{0}}\frac{f(x+h).g(x+h) – f(x).g(x)}{h}}$$

Now, we’ll have to do some manipulation in order to proceed conveniently. Let us add and subtract f(x + h).g(x) in the numerator. We’ll get: $${ \frac{d}{dx}(f(x)g(x)) = Lim_{h\rightarrow{0}}\frac{f(x+h).g(x+h) – f(x).g(x) + f(x+h).g(x) – f(x+h).g(x)}{h}}$$

Rearranging the right-hand side,
$${R.H.S. = Lim_{h\rightarrow{0}}\frac{f(x+h)[.g(x+h) – g(x)] + g(x)[f(x+h) – f(x)]}{h}}$$
$${Lim_{h\rightarrow{0}}\frac{f(x+h)[.g(x+h) – g(x)]}{h} + Lim_{h\rightarrow{0}}\frac{g(x)[f(x+h) – f(x)]}{h}}$$

Now, you must remember the product rule of limits i.e.$${ Lim_{x\rightarrow{0}}[f(x).g(x)] =  [Lim_{x\rightarrow{0}}f(x)].[Lim_{x\rightarrow{0}}g(x)] }$$

Applying this rule to the equation above, we can get:
$${ (Lim_{h\rightarrow{0}}f(x+h)).(Lim_{h\rightarrow{0}}\frac{[g(x+h) – g(x)]}{h}) + (Lim_{h\rightarrow{0}}g(x))(Lim_{h\rightarrow{0}}\frac{[f(x+h) – f(x)]}{h}})$$

Work out the individual limits and use the definition of the derivative to get:
$${ f(x)\frac{d}{dx}g(x) + g(x)\frac{d}{dx}f(x)}$$

This concludes the proof of the Product Rule.

Theorem 4: The derivative of the quotient of two functions is given by the Quotient Rule.

$${\frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{[\frac{d}{dx}f(x)].g(x) – [\frac{d}{dx}g(x)].f(x)}{g(x)^2}} $$

Proof: Let us begin from the definition of the derivative of the function \({\frac{f(x)}{g(x)}}\)
$${ \frac{d}{dx}(\frac{f(x)}{g(x)}) = Lim_{h\rightarrow{0}}\frac{\frac{f(x+h)}{g(x+h)} – \frac{f(x)}{g(x)}}{h}}$$
or
$${ \frac{d}{dx}(\frac{f(x)}{g(x)}) = Lim_{h\rightarrow{0}}\frac{1}{h}\frac{f(x+h)g(x) – f(x)g(x+h)}{g(x+h)g(x)}}$$

Now we must manipulate the equations to conform to our proof. Let us add and subtract f(x).g(x) in the numerator on the right-hand side. We’ll get something like:
$${ R.H.S. = Lim_{h\rightarrow{0}}\frac{1}{h}\frac{f(x+h)g(x) – f(x)g(x+h) + f(x).g(x) – f(x).g(x)}{g(x+h)g(x)}}$$
$${ Lim_{h\rightarrow{0}}\frac{1}{g(x+h)g(x)}\frac{f(x+h)g(x) – f(x)g(x+h) + f(x).g(x) – f(x).g(x)}{h}}$$

Rearranging the numerator and subsequently, breaking it up into fractions, we can get:
$${Lim_{h\rightarrow{0}}\frac{1}{g(x+h)g(x)}\frac{g(x)[(f(x+h) – f(x)] – f(x)[g(x+h) – g(x)] }{h}}$$
$${ Lim_{h\rightarrow{0}}\frac{1}{g(x+h)g(x)}(g(x)\frac{(f(x+h) – f(x)}{h} – f(x)\frac{g(x+h) – g(x)}{h})}$$

Using the properties of limits again, we can split it over the numerator and the denominator as:
$${ \frac{1}{Lim_{h\rightarrow{0}}(g(x+h))Lim_{h\rightarrow{0}}(g(x))}((Lim_{h\rightarrow{0}}(g(x))Lim_{h\rightarrow{0}}[\frac{(f(x+h) – f(x)}{h}] \\ – (Lim_{h\rightarrow{0}}f(x))Lim_{h\rightarrow{0}}[\frac{g(x+h) – g(x)}{h}])}$$

Working out the individual limits and using the definition of the derivative, one can get:
$${ \frac{[\frac{d}{dx}f(x)].g(x) – [\frac{d}{dx}g(x)].f(x)}{g(x)^2} }$$

This proves the Quotient Rule.

Theorem 5: The derivative of a constant is equal to 0

$$ {{\frac{d}{dx}(c) = 0}\text{ (where c is a constant)}} $$

It is very straightforward to prove it from the definition of the derivative itself.

The Power Rule

$$ {\frac{d}{dx}(x^n) = nx^{n-1} \text{ (where n is a real number) }}$$

It can be proved using various ways if necessary. You could check it from other sources if you need the proof of this rule. But as of now, just work with it!

Learn Algebra of Continous Functions here. 

Solved Examples for You

Question: Find the derivative of the function \({ f(x) = \frac{sin x}{x} }\).

Solution: We can compute such derivatives using both, the Product and the Quotient Rule. We’ll show you these methods one by one.

Case 1: Using the product rule, consider the given function as \( { f(x) = (sin x).(x^{-1})}\).

Then its derivative can be computed as:
$${ \frac{d}{dx}f(x) = (\frac{d}{dx}sin x).(x^{-1}) + (\frac{d}{dx}{x^{-1}}).sin x }$$
Use cos x as the derivative of sin x, and the power rule to differentiate the second function; and arrive at –
$${ \frac{d}{dx}f(x) = \frac{cos x}{x} + (\frac{-1}{x^2}).sin x }$$
$${ \frac{d}{dx}f(x) = \frac{cos x}{x} – \frac{sin x}{x^2} }$$
which is the required result.

Case 2: Using the quotient rule, consider the given function as \( { f(x) = \frac{sin x}{x}}\).

Then its derivative can be computed as:
$${ \frac{d}{dx}f(x) = \frac{ (\frac{d}{dx}sin x).x – (\frac{d}{dx}{x}).sin x}{x^2} }$$
Again, use cos x as the derivative of sin x, and the power rule to differentiate the second function; and arrive at –
$${ \frac{d}{dx}f(x) = \frac{(cos x).x – 1.(sin x)}{x^2}}$$
$${ \frac{d}{dx}f(x) = { \frac{cos x}{x} – \frac{sin x}{x^2}} }$$

which is the same result as the one we got by applying the Product Rule. Thus, you may apply any of the rules discussed above as per your convenience.

This concludes our discussion on this topic.

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