Rules of Differentiation

Differentiation follows a set of basic rules that help us find derivatives easily. These rules allow us to differentiate different types of functions without always using the limit definition.

By using these rules along with the derivatives of simple functions, we can differentiate more complex functions that involve addition, multiplication, division, and compositions of basic functions. These rules make the process of differentiation simpler and more efficient.

Constant Multiplication Rule:

\[ \frac{d}{dx} \left[ k f(x) \right] = k \frac{d}{dx} f(x) \]

Proof: Using the first principles definition,

\[ \frac{d}{dx} \left[ k f(x) \right] = \lim_{\Delta x \to 0} \frac{k f(x + \Delta x) - k f(x)}{\Delta x}. \]

Factoring out \( k \),

\[ = k \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}. \]

Recognizing the definition of the derivative,

\[ = k \frac{d}{dx} f(x). \]

Sum Rule:

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

Proof: Using the first principles definition,

\[ \frac{d}{dx} [f(x) + g(x)] = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) + g(x + \Delta x) - f(x) - g(x)}{\Delta x}. \]

Splitting the fraction,

\[ = \lim_{\Delta x \to 0} \left[ \frac{f(x + \Delta x) - f(x)}{\Delta x} + \frac{g(x + \Delta x) - g(x)}{\Delta x} \right]. \]

Taking the limit separately,

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

Example

Now we can differentiate a function like \( 3e^x + 4\ln x - \cos x \). Using the sum and constant multiple rules,

\[ \frac{d}{dx} (3e^x + 4\ln x - \cos x) \]

Applying the differentiation rule separately to each term,

\[ = \frac{d}{dx} (3e^x) + \frac{d}{dx} (4\ln x) - \frac{d}{dx} (\cos x). \]

Using the constant multiple rule,

\[ = 3 \frac{d}{dx} e^x + 4 \frac{d}{dx} \ln x - \frac{d}{dx} \cos x. \]

Substituting the known derivatives,

\[ = 3e^x + 4 \cdot \frac{1}{x} + \sin x. \]

Simplifying,

\[ = 3e^x + \frac{4}{x} + \sin x. \]

Product Rule

The product rule states that if \( f(x) \) and \( g(x) \) are both differentiable functions, then the derivative of their product is given by

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

This rule allows us to differentiate functions that are products of two differentiable functions without expanding them first.

Proof Using First Principles

By definition, the derivative is

\[ \frac{d}{dx} [f(x) g(x)] = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) g(x + \Delta x) - f(x) g(x)}{\Delta x}. \]

Adding and subtracting \( f(x) g(x + \Delta x) \),

\[ = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) g(x + \Delta x) - f(x) g(x + \Delta x) + f(x) g(x + \Delta x) - f(x) g(x)}{\Delta x}. \]

Rearranging terms,

\[ = \lim_{\Delta x \to 0} \left[ \frac{f(x + \Delta x) - f(x)}{\Delta x} g(x + \Delta x) + f(x) \frac{g(x + \Delta x) - g(x)}{\Delta x} \right]. \]

Taking the limit,

\[ = \left( \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \right) g(x) + f(x) \left( \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} \right). \]

Recognizing these as \( f'(x) \) and \( g'(x) \),

\[ = f'(x) g(x) + f(x) g'(x). \]

Thus, the product rule is proven:

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

Example

Now we can differentiate a function like \( 2x e^x \). Using the constant multiple rule, we take \( 2 \) out first:

\[ \frac{d}{dx} (2x e^x) = 2 \frac{d}{dx} (x e^x). \]

Applying the product rule to \( x e^x \), where \( f(x) = x \) and \( g(x) = e^x \),

\[ \frac{d}{dx} (x e^x) = x \frac{d}{dx} e^x + e^x \frac{d}{dx} x. \]

Computing the derivatives,

\[ = x e^x + e^x. \]

Thus,

\[ \frac{d}{dx} (2x e^x) = 2 (x e^x + e^x). \]

Simplifying,

\[ = 2 e^x (x + 1). \]

Example

Differentiate \( f(x) = x \ln x - x^3 + 2x + 4 \).

Solution:

\[ \frac{d}{dx} (x \ln x - x^3 + 2x + 4) = \frac{d}{dx} (x \ln x) - \frac{d}{dx} (x^3) + \frac{d}{dx} (2x) + \frac{d}{dx} (4). \]

Applying the product rule to \( x \ln x \),

\[ \frac{d}{dx} (x \ln x) = x \frac{d}{dx} (\ln x) + \ln x \frac{d}{dx} (x) = x \cdot \frac{1}{x} + \ln x \cdot 1 = 1 + \ln x. \]

Computing other derivatives,

\[ \frac{d}{dx} (x^3) = 3x^2, \quad \frac{d}{dx} (2x) = 2, \quad \frac{d}{dx} (4) = 0. \]

Thus,

\[ \frac{d}{dx} (x \ln x - x^3 + 2x + 4) = (1 + \ln x) - 3x^2 + 2. \]

Final Answer:

\[ \ln x - 3x^2 + 3. \]

Example

Differentiate \( f(x) = e^x (x + \ln x) \).

Solution:

Applying the product rule,

\[ \frac{d}{dx} \left( e^x (x + \ln x) \right) = e^x \frac{d}{dx} (x + \ln x) + (x + \ln x) \frac{d}{dx} e^x. \]

Now differentiating each term,

\[ \frac{d}{dx} (x + \ln x) = \frac{d}{dx} (x) + \frac{d}{dx} (\ln x) = 1 + \frac{1}{x}, \]

\[ \frac{d}{dx} e^x = e^x. \]

Substituting back,

\[ \frac{d}{dx} \left( e^x (x + \ln x) \right) = e^x \left( 1 + \frac{1}{x} \right) + (x + \ln x) e^x. \]

Final Answer:

\[ e^x \left( 1 + \frac{1}{x} + x + \ln x \right). \]

Quotient Rule

The quotient rule states that if \( f(x) \) and \( g(x) \) are differentiable functions, then the derivative of their quotient is given by

\[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{g(x) \frac{d}{dx} f(x) - f(x) \frac{d}{dx} g(x)}{[g(x)]^2}. \]

This rule is essential for differentiating functions where one function is divided by another.

Proof Using First Principles

By definition, the derivative is

\[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \lim_{\Delta x \to 0} \frac{\frac{f(x + \Delta x)}{g(x + \Delta x)} - \frac{f(x)}{g(x)}}{\Delta x}. \]

Rewriting the numerator as a single fraction,

\[ \frac{f(x + \Delta x)}{g(x + \Delta x)} - \frac{f(x)}{g(x)} = \frac{f(x + \Delta x) g(x) - f(x) g(x + \Delta x)}{g(x + \Delta x) g(x)}. \]

Thus,

\[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) g(x) - f(x) g(x + \Delta x)}{\Delta x} \cdot \frac{1}{g(x + \Delta x) g(x)}. \]

Rewriting the numerator,

\[ f(x + \Delta x) g(x) - f(x) g(x + \Delta x) = \left( f(x + \Delta x) g(x) - f(x) g(x) \right) + \left( f(x) g(x) - f(x) g(x + \Delta x) \right). \]

Factoring each term,

\[ = g(x) \left( f(x + \Delta x) - f(x) \right) - f(x) \left( g(x + \Delta x) - g(x) \right). \]

Substituting back,

\[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \lim_{\Delta x \to 0} \frac{g(x) (f(x + \Delta x) - f(x)) - f(x) (g(x + \Delta x) - g(x))}{\Delta x} \cdot \frac{1}{g(x + \Delta x) g(x)}. \]

Taking the limit,

\[ = \frac{g(x) f'(x) - f(x) g'(x)}{[g(x)]^2}. \]

Thus, the quotient rule is proven:

\[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{g(x) f'(x) - f(x) g'(x)}{[g(x)]^2}. \]

Example

Differentiate \( f(x) = \frac{\ln x}{x} \)

Solution:

Let \( f(x) = \frac{\ln x}{x} \), where
\( f(x) = \ln x \) and \( g(x) = x \).

Applying the quotient rule,

\[ \frac{d}{dx} \left[ \frac{\ln x}{x} \right] = \frac{x \frac{d}{dx} (\ln x) - \ln x \frac{d}{dx} (x)}{x^2}. \]

Differentiating each term,

\[ \frac{d}{dx} (\ln x) = \frac{1}{x}, \quad \frac{d}{dx} (x) = 1. \]

Substituting these,

\[ = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2}. \]

Simplifying,

\[ = \frac{1 - \ln x}{x^2}. \]

Final Answer:

\[ \frac{1 - \ln x}{x^2}. \]

Example

Differentiate \( f(x) = \frac{\sin x}{x^2 + 1} \)

Solution:

Let
\( f(x) = \sin x \),
\( g(x) = x^2 + 1 \).

Applying the quotient rule,

\[ \frac{d}{dx} \left[ \frac{\sin x}{x^2 + 1} \right] = \frac{(x^2 + 1) \frac{d}{dx} (\sin x) - \sin x \frac{d}{dx} (x^2 + 1)}{(x^2 + 1)^2}. \]

Computing the derivatives,

\[ \frac{d}{dx} (\sin x) = \cos x, \quad \frac{d}{dx} (x^2 + 1) = 2x. \]

Substituting these,

\[ = \frac{(x^2 + 1) \cos x - \sin x \cdot 2x}{(x^2 + 1)^2}. \]

Final Answer:

\[ \frac{(x^2 + 1) \cos x - 2x \sin x}{(x^2 + 1)^2}. \]

Corollary: Reciprocal Rule

The reciprocal rule states that if \( f(x) \) is a differentiable function and \( f(x) \neq 0 \), then the derivative of its reciprocal is given by

\[ \frac{d}{dx} \left[ \frac{1}{f(x)} \right] = -\frac{f'(x)}{[f(x)]^2}. \]

This follows directly from the quotient rule by setting \( g(x) = 1 \), so that

\[ \frac{d}{dx} \left[ \frac{1}{f(x)} \right] = \frac{f(x) \cdot \frac{d}{dx} (1) - 1 \cdot \frac{d}{dx} f(x)}{[f(x)]^2}. \]

Since \( \frac{d}{dx} (1) = 0 \), this simplifies to

\[ \frac{d}{dx} \left[ \frac{1}{f(x)} \right] = -\frac{f'(x)}{[f(x)]^2}. \]

Thus, the reciprocal rule provides a quick way to differentiate functions of the form \( 1/f(x) \) without applying the full quotient rule explicitly.

Example

Differentiate \( f(x) = \frac{1}{x^2 + x + 1} \) using the reciprocal rule.

Solution:

Using the reciprocal rule:

\[ \frac{d}{dx} \left[ \frac{1}{f(x)} \right] = -\frac{f'(x)}{[f(x)]^2}. \]

Let \( f(x) = x^2 + x + 1 \), then

\[ \frac{d}{dx} (x^2 + x + 1) = 2x + 1. \]

Applying the reciprocal rule,

\[ \frac{d}{dx} \left[ \frac{1}{x^2 + x + 1} \right] = -\frac{2x + 1}{(x^2 + x + 1)^2}. \]

Final Answer:

\[ -\frac{2x + 1}{(x^2 + x + 1)^2}. \]

Example

Differentiate \( f(x) = \frac{5}{\sin x + \cos x} \) using the reciprocal rule.

Solution:

Using the reciprocal rule,

\[ \frac{d}{dx} \left[ \frac{5}{g(x)} \right] = -\frac{5 g'(x)}{[g(x)]^2}. \]

Let \( g(x) = \sin x + \cos x \), then

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

Applying the reciprocal rule,

\[ \frac{d}{dx} \left[ \frac{5}{\sin x + \cos x} \right] = -\frac{5 (\cos x - \sin x)}{(\sin x + \cos x)^2}. \]

Final Answer:

\[ -\frac{5 (\cos x - \sin x)}{(\sin x + \cos x)^2}. \]

Chain Rule (Composition Rule)

For a composition of two differentiable functions \( f(g(x)) \), the derivative is given by

\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x). \]

This rule allows us to differentiate composite functions efficiently by first differentiating the outer function and then multiplying by the derivative of the inner function.

Proof Using First Principles*

By definition of the derivative,

\[ \frac{d}{dx} f(g(x)) = \lim_{\Delta x \to 0} \frac{f(g(x + \Delta x)) - f(g(x))}{\Delta x}. \]

Multiplying and dividing by \( g(x + \Delta x) - g(x) \),

\[ = \lim_{\Delta x \to 0} \left[ \frac{f(g(x + \Delta x)) - f(g(x))}{g(x + \Delta x) - g(x)} \cdot \frac{g(x + \Delta x) - g(x)}{\Delta x} \right]. \]

Taking limits separately,

\[ \lim_{\Delta x \to 0} \frac{f(g(x + \Delta x)) - f(g(x))}{g(x + \Delta x) - g(x)} \quad \text{and} \quad \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x}. \]

For the first limit, let

\[ u = g(x), \quad u + \Delta u = g(x + \Delta x), \]

so that \( \Delta u = g(x + \Delta x) - g(x) \). Then the first limit transforms into

\[ \lim_{\Delta u \to 0} \frac{f(u + \Delta u) - f(u)}{\Delta u}. \]

By definition, this is \( f'(u) \), and since \( u = g(x) \), we obtain

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

For the second limit,

\[ \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} = g'(x). \]

Thus, multiplying both limits together,

\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x). \]

\(\blacksquare\)

---

From the proof, we can see something important: the composition \( y = f(g(x)) \) can be understood as a chain of transformations:

\[ x \quad \xrightarrow{g} \quad g(x) \quad \xrightarrow{f} \quad f(g(x)). \]

Here, \( x \) is first mapped to \( g(x) \), and then \( g(x) \) is mapped to \( f(g(x)) \). The function \( g(x) \) acts as an intermediate output in this process.

If we introduce a new variable \( u \) such that

\[ u = g(x), \]

then differentiating both sides gives

\[ \frac{du}{dx} = g'(x). \]

Since \( y = f(u) = f(g(x)) \), differentiating with respect to \( x \),

\[ \frac{dy}{dx} = \frac{d}{dx} f(g(x)) = f'(g(x)) g'(x). \]

Using the intermediate variable \( u \), this can be rewritten as

\[ \frac{dy}{dx} = f'(u) \cdot g'(x). \]

Recognizing that \( f'(u) = \frac{dy}{du} \) and \( g'(x) = \frac{du}{dx} \), we obtain

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}. \]

Thus, the derivative of a composite function can be written in the chain rule form:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}. \]

This formulation highlights that the total rate of change of \( y \) with respect to \( x \) can be computed by first differentiating \( y \) with respect to the intermediate variable \( u \) and then multiplying by the derivative of \( u \) with respect to \( x \).

Example

Differentiate \( y = \sin(\sin x) \).

Solution:
Let

\[ y = \sin(\sin x). \]

Defining \( u = \sin x \), we rewrite \( y \) as

\[ y = \sin u. \]

Differentiating both sides using the chain rule,

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}. \]

Since \( \frac{dy}{du} = \cos u \) and \( \frac{du}{dx} = \cos x \), we substitute,

\[ \frac{dy}{dx} = \cos u \cdot \cos x. \]

Substituting \( u = \sin x \),

\[ \frac{dy}{dx} = \cos (\sin x) \cos x. \]

To do this quickly, first differentiate the outermost function in the composition with respect to what is inside, treating it mentally as a single object, say \( u \). Then multiply by the derivative of \( u \) with respect to \( x \).

Final Answer:

\[ \frac{dy}{dx} = \cos (\sin x) \cos x. \]

Example

Differentiate \( f(x) = \ln (\sin x) \).

Solution:

To compute this derivative efficiently, first differentiate the outermost function, \( \ln (\cdot) \), with respect to what is inside, treating \( \sin x \) as a single object. The derivative of \( \ln (\cdot) \) is \( \frac{1}{\text{(inside function)}} \), so

\[ \frac{d}{dx} \ln (\sin x) = \frac{1}{\sin x} \cdot \frac{d}{dx} (\sin x). \]

Now, differentiate \( \sin x \),

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

Thus,

\[ \frac{d}{dx} \ln (\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x. \]

Final Answer:

\[ \frac{d}{dx} \ln (\sin x) = \cot x. \]

Chain Rule of Differentiation for Multiple Compositions

For a function involving multiple compositions, such as \( f(g(h(k(x)))) \), the differentiation process extends by repeatedly applying the chain rule:

\[ \frac{d}{dx} f(g(h(k(x)))) = f'(g(h(k(x)))) \cdot g'(h(k(x))) \cdot h'(k(x)) \cdot k'(x). \]

To compute this efficiently, first differentiate the outermost function \( f(\cdot) \) with respect to its argument, treating everything inside as a single object. This gives \( f'(g(h(k(x)))) \). Then, multiply by the derivative of the next inner function, \( g(h(k(x))) \), again treating \( h(k(x)) \) as a single object. Continue this process until reaching the innermost function \( k(x) \).

Thus, differentiation proceeds step by step, always applying the derivative to the outermost function first and then moving inward. This method ensures that complex compositions can be differentiated efficiently and systematically without explicitly substituting intermediate variables.

The expression

\[ \frac{dy}{du} \cdot \frac{du}{dx} = \frac{dy}{dx} \]

suggests a formal property of derivatives that resembles cancellation of the intermediate variable \( u \). However, such cancellation is not literal but rather a consequence of the chain rule, which states that if a function \( y \) depends on \( u \), and \( u \) depends on \( x \), then \( y \) can be expressed as a function of \( x \), and its derivative satisfies

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}. \]

The notation

\[ \frac{dy}{du}, \quad \frac{du}{dx}, \quad \text{and} \quad \frac{dy}{dx} \]

suggests a fraction-like behavior, but these expressions do not represent actual fractions; rather, they are limit-based definitions of derivatives. Specifically, writing

\[ \frac{dy}{du} = \lim_{\Delta u \to 0} \frac{\Delta y}{\Delta u}, \quad \frac{du}{dx} = \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} \]

and formally substituting into their product gives

\[ \frac{dy}{du} \cdot \frac{du}{dx} = \left( \lim_{\Delta u \to 0} \frac{\Delta y}{\Delta u} \right) \cdot \left( \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} \right). \]

Since \( \Delta u \) appears in both fractions, one can rewrite

\[ \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x} = \frac{\Delta y}{\Delta x}, \]

and taking the limit as \( \Delta x \to 0 \) recovers

\[ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}. \]

This justifies the chain rule rigorously. Although the cancellation of \( du \) is only symbolic, the notation is useful as a mnemonic device for applying the chain rule efficiently.

Example

Differentiate \( f(x) = \sin(\cos(\ln x)) \).

Solution:

Differentiate the outermost function \( \sin(\cos(\ln x)) \) first:

\[ \frac{d}{dx} \sin(\cos(\ln x)) = \cos(\cos(\ln x)) \cdot \frac{d}{dx} \cos(\ln x). \]

Differentiate \( \cos(\ln x) \):

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

Since \( \frac{d}{dx} \ln x = \frac{1}{x} \),

\[ \frac{d}{dx} \cos(\ln x) = -\sin(\ln x) \cdot \frac{1}{x}. \]

Substituting this into the previous expression:

\[ \frac{d}{dx} \sin(\cos(\ln x)) = \cos(\cos(\ln x)) \cdot \left(-\sin(\ln x) \cdot \frac{1}{x} \right). \]

Thus,

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

Example

Differentiate \( f(x) = \sin(\cos(\ln x)) \).

Solution:

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

Differentiating the outermost function first, treating everything inside as a single entity,

\[ \frac{d}{dx} \sin(\cos(\ln x)) = \cos(\cos(\ln x)) \cdot \frac{d}{dx} \cos(\ln x). \]

The derivative of \( \sin u \) is \( \cos u \), and the argument remains unchanged, multiplying by the derivative of the inside function \( \cos(\ln x) \). Now, differentiating \( \cos(\ln x) \),

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

The derivative of \( \cos v \) is \( -\sin v \), keeping \( \ln x \) unchanged, and multiplying by the derivative of \( \ln x \). Since

\[ \frac{d}{dx} \ln x = \frac{1}{x}, \]

substituting this,

\[ \frac{d}{dx} \sin(\cos(\ln x)) = \cos(\cos(\ln x)) \cdot [-\sin(\ln x)] \cdot \frac{1}{x}. \]

This simplifies to

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

Thus,

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

Each differentiation step follows the chain rule, treating the inner function as a single entity and differentiating from the outside in. This ensures that the composition is handled systematically without introducing unnecessary intermediate variables.

Example

Differentiate \( f(x) = \sqrt{1 + x^2} \).

Solution:

\[ \frac{d}{dx} \sqrt{1 + x^2} \]

Differentiating the outermost function first, treating \( 1 + x^2 \) as a single entity,

\[ = \frac{d}{dx} (1 + x^2)^{\frac{1}{2}} = \frac{1}{2} (1 + x^2)^{-\frac{1}{2}} \cdot \frac{d}{dx} (1 + x^2). \]

The derivative of \( u^{\frac{1}{2}} \) is \( \frac{1}{2} u^{-\frac{1}{2}} \), keeping \( 1 + x^2 \) unchanged, and multiplying by the derivative of \( 1 + x^2 \). Now, differentiating \( 1 + x^2 \),

\[ = \frac{d}{dx} \sqrt{1 + x^2} = \frac{1}{2} (1 + x^2)^{-\frac{1}{2}} \cdot (2x). \]

Since the derivative of \( x^2 \) is \( 2x \) and the derivative of the constant \( 1 \) is zero, we substitute this result. Simplifying the expression,

\[ = \frac{d}{dx} \sqrt{1 + x^2} = \frac{2x}{2(1 + x^2)^{\frac{1}{2}}}. \]

Canceling the factor of 2,

\[ = \frac{d}{dx} \sqrt{1 + x^2} = \frac{x}{\sqrt{1 + x^2}}. \]

Thus,

\[ \frac{d}{dx} f(x) = \frac{x}{\sqrt{1 + x^2}}. \]

Each differentiation step applies the chain rule systematically, ensuring that the composition is handled efficiently while maintaining clarity.

Example

I. Differentiate \( f(x) = x - \sqrt{x^2 - 1} \).

Solution:

\[ \frac{d}{dx} \left( x - \sqrt{x^2 - 1} \right) \]

Differentiating term by term,

\[ = \frac{d}{dx} x - \frac{d}{dx} \sqrt{x^2 - 1}. \]

Since \( \frac{d}{dx} x = 1 \),

\[ = 1 - \frac{d}{dx} (x^2 - 1)^{\frac{1}{2}}. \]

Differentiating the outermost function first, treating \( x^2 - 1 \) as a single entity,

\[ = 1 - \frac{1}{2} (x^2 - 1)^{-\frac{1}{2}} \cdot \frac{d}{dx} (x^2 - 1). \]

The derivative of \( x^2 - 1 \) is \( 2x \), so,

\[ = 1 - \frac{1}{2} (x^2 - 1)^{-\frac{1}{2}} \cdot (2x). \]

Canceling the factor of 2,

\[ = 1 - \frac{x}{\sqrt{x^2 - 1}}. \]

Thus,

\[ \frac{d}{dx} f(x) = 1 - \frac{x}{\sqrt{x^2 - 1}}. \]

Each differentiation step follows the chain rule systematically, treating the inner function as a single entity before differentiating it explicitly. \(\blacksquare\)

II. Differentiate \( f(x) = \ln \left( x - \sqrt{x^2 - 1} \right) \).

Solution:

\[ \frac{d}{dx} \ln \left( x - \sqrt{x^2 - 1} \right) \]

Using the derivative of \( \ln u \),

\[ = \frac{1}{x - \sqrt{x^2 - 1}} \cdot \frac{d}{dx} \left( x - \sqrt{x^2 - 1} \right). \]

Differentiating \( x - \sqrt{x^2 - 1} \),

\[ = \frac{1}{x - \sqrt{x^2 - 1}} \cdot \left( 1 - \frac{d}{dx} \sqrt{x^2 - 1} \right). \]

Differentiating \( \sqrt{x^2 - 1} \), treating \( x^2 - 1 \) as a single entity,

\[ = \frac{1}{x - \sqrt{x^2 - 1}} \cdot \left( 1 - \frac{1}{2} (x^2 - 1)^{-\frac{1}{2}} \cdot \frac{d}{dx} (x^2 - 1) \right). \]

Since \( \frac{d}{dx} (x^2 - 1) = 2x \),

\[ = \frac{1}{x - \sqrt{x^2 - 1}} \cdot \left( 1 - \frac{1}{2} (x^2 - 1)^{-\frac{1}{2}} \cdot 2x \right). \]

Canceling the factor of 2,

\[ = \frac{1}{x - \sqrt{x^2 - 1}} \cdot \left( 1 - \frac{x}{\sqrt{x^2 - 1}} \right). \]

Rewriting the fraction,

\[ = \frac{1 - \frac{x}{\sqrt{x^2 - 1}}}{x - \sqrt{x^2 - 1}}. \]

Multiplying numerator and denominator by \( \sqrt{x^2 - 1} \) to simplify,

\[ = \frac{\sqrt{x^2 - 1} - x}{(x - \sqrt{x^2 - 1}) \sqrt{x^2 - 1}}. \]

Since \( \sqrt{x^2 - 1} - x = -(x - \sqrt{x^2 - 1}) \),

\[ = \frac{-(x - \sqrt{x^2 - 1})}{(x - \sqrt{x^2 - 1}) \sqrt{x^2 - 1}}. \]

Canceling \( (x - \sqrt{x^2 - 1}) \) from numerator and denominator,

\[ = -\frac{1}{\sqrt{x^2 - 1}}. \]

Thus,

\[ \frac{d}{dx} f(x) = -\frac{1}{\sqrt{x^2 - 1}}. \]