Differentiation

In this page, we will come across proofs for some rules of differentiation which we use for most differentiation problems. In proving these rules, the standard "PEMDAS" (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) will be used.

Let \(a\) be an arbitrary real number. The constant rule states that

\[ \frac{d\ (a)}{dx} = 0. \]

Proof:

Recall that for an arbitrary function \(f(x)\),

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

Let \(f(x) = a\). Substitute \(x+h\) for \(x\). Since \(f(x) = a\) is a constant function, \(f(x+h) = a\).

\[ \frac{d\ (a)}{dx} = \lim_{h\ \to\ 0} \frac{a - a}{h} = \lim_{h\ \to\ 0} \frac{0}{h} = \lim_{h\ \to\ 0} 0 = 0.\ _\square \]

Let \(a\) be an arbitrary real number, and \(g(x)\) an arbitrary differentiable function.

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

Proof:

Recall that for an arbitrary function \(f(x)\),

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

Let \(f(x) = a \cdot g(x)\). Substitute \(x+h\) for \(x\) to get \(f(x+h) = a \cdot g(x+h)\).

\[\begin{align} \frac{d\ \big(a \cdot g(x)\big)}{dx} &= \lim_{h\ \to\ 0} \frac{a \cdot g(x+h) - a \cdot g(x)}{h} \\ &= \lim_{h\ \to\ 0} \frac{a \big( g(x+h) - g(x) \big)}{h} \\ &= \left(\lim_{h\ \to\ 0} a\right) \lim_{h\ \to\ 0} \frac{g(x+h) - g(x)}{h}. \end{align}\]

By definitions of limit and derivative,

\[ \frac{d\ \big(a \cdot g(x)\big)}{dx} = a \cdot g'(x).\ _\square\]

Let \(n\) and \(x\) be real numbers, with \(n \neq 0\) and \(x \neq 0\).

The power rule states that

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

Proof 1:

Recall that for an arbitrary function \(f(x)\),

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

Let \(f(x) = x^n\). Substitute \(x+h\) for \(x\) to get \(f(x+h) = (x+h)^n\).

\[ \frac{d\ (x^n)}{dx} = \lim_{h\ \to\ 0} \frac{(x+h)^n - x^n}{h}. \]

The binomial theorem states that

\[\begin{align} (x+h)^n &= \sum\limits_{c\ =\ 0}^n\ _nC_c\ x^{n-c}\ h^c\\ &=\ _nC_0\ x^{n-0}\ h^0 +\ _nC_1\ x^{n-1}\ h^1 +\ _nC_2\ x^{n-2}\ h^2 +\cdots+\ _nC_{n-1}\ x^1\ h^{n-1} +\ _nC_n\ x^{n-n}\ h^n \\ &= x^n + nx^{n-1}h + \frac{n(n-1)}{2} x^{n-2}h^2 +\cdots + h^n. \end{align}\]

Substitute the value of \((x+h)^n\) to get

\[ \frac{d\ (x^n)}{dx} = \lim_{h\ \to\ 0} \frac{\left(x^n + nx^{n-1}h + \frac{n(n-1)}{2} x^{n-2}h^2 +\cdots + h^n\right) - x^n}{h}. \]

Cancel \(x^n\) and \(h\) to get

\[\begin{align} \frac{d\ (x^n)}{dx} &= \lim_{h\ \to\ 0} \frac{\left(nx^{n-1}h + \frac{n(n-1)}{2} x^{n-2}h^2 +\cdots + h^n\right)}{h} \\\\ &= \lim_{h\ \to\ 0} \frac{h\ \left(nx^{n-1} + \frac{n(n-1)}{2} x^{n-2}h +\cdots + h^{n-1}\right)}{h} \\\\ &= \lim_{h\ \to\ 0} \left(nx^{n-1} + \frac{n(n-1)}{2} x^{n-2}h +\cdots + h^{n-1}\right). \end{align}\]

By definition of limit,

\[ \frac{d\ (x^n)}{dx} = nx^{n-1} + 0 +\cdots + 0 = \frac{d\ (x^n)}{dx} = nx^{n-1}.\ _\square\]

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Let \(n\) and \(x\) be real numbers, with \(n \neq 0\) and \(x \neq 0\).

As per power rule for differentiation, we know that

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

(The Greek word "\(\delta\)" means "a small change in.")

Proof 2:

Let \(y = x^n\).

We will look at what kind of small change we will get in \(y\) if we change \(x\) by adding a small amount of \(x\) (denoted by \(\delta x\)):

\[ y = x^n \implies y + \delta y = (x + \delta x)^n. \]

Expand \((x + \delta x) ^n\) using binomial expansion to get

\[ y + \delta y = x^n + n \cdot x^{n-1}(\delta x) + \frac{n(n-1) \cdot x^{n-2}(\delta x)^2}{2!} + (\text{terms with higher ascending powers of }\delta x). \]

Substitute \(x^n\) for \(y\) and divide both sides of the equation by \(\delta x\) to get

\[\begin{align} x^n + \delta y &= x^n + n \cdot x^{n-1}(\delta x) + \frac{n(n-1) \cdot x^{n-2}(\delta x)^2}{2!} + (\text{terms with higher ascending powers of }\delta x) \\ \\ \frac{\delta y}{\delta x} &= n \cdot x^{n-1}+ \frac{n(n-1) \cdot x^{n-2}(\delta x)}{2!} + (\text{terms with higher ascending powers of }\delta x). \end{align}\]

As \(\delta x \to 0,\) all terms with powers of \(\delta x\) that are greater than \(0\) become so small that they can be ignored:

\[ \displaystyle \lim_{\delta x \to 0} \frac{\delta y}{\delta x} = n \cdot x^{n-1}.\ _\square\]

Let \(x\) be an arbitrary variable, then

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

Proof:

Let \(n\) and \(x\) be real numbers, with \(n \neq 0\) and \(x \neq 0\).

The power rule states that

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

Let \(x^n = x = x^1\).

\[ \frac{d\ (x)}{dx} = 1 \cdot x^{1-1} = x^0 = 1.\ _\square\]

Let \(f(x)\) and \(g(x)\) be arbitrary differentiable functions.

Recall that for an arbitrary function \(a(x)\),

\[ \frac{d\ \big(a(x)\big)}{dx} = a'(x) = \lim_{h\ \to\ 0} \frac{a(x+h) - a(x)}{h}. \]

The sum rule states that

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

Proof:

Let \(a(x) = f(x) + g(x)\). Substitute \(x+h\) for \(x\) to get \(a(x+h) = f(x+h) + g(x+h)\):

\[ \begin{align} \frac{d\ \big(f(x) + g(x)\big)}{dx} &= \lim_{h\ \to\ 0} \frac{f(x+h) + g(x+h) - \big(f(x) + g(x)\big)}{h} \\ &= \lim_{h\ \to\ 0} \frac{f(x+h) + g(x+h) - f(x) - g(x)}{h} \\ &= \lim_{h\ \to\ 0} \frac{f(x+h) - f(x)}{h} + \lim_{h\ \to\ 0} \frac{g(x+h) - g(x)}{h}. \end{align}\]

By definition of derivative,

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

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The difference rule states that

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

Proof:

Let \(a(x) = f(x) - g(x)\). Substitute \(x+h\) for \(x\) to get \(a(x+h) = f(x+h) - g(x+h)\):

\[ \begin{align} \frac{d\ \big(f(x) - g(x)\big)}{dx} &= \lim_{h\ \to\ 0} \frac{f(x+h) - g(x+h) - \big(f(x) - g(x)\big)}{h} \\ &= \lim_{h\ \to\ 0} \frac{f(x+h) - g(x+h) - f(x) + g(x)}{h} \\ &= \lim_{h\ \to\ 0} \frac{f(x+h) - f(x)}{h} - \lim_{h\ \to\ 0} \frac{g(x+h) - g(x)}{h}. \end{align}\]

By definition of derivative,

\[ \frac{d\ \big(f(x) - g(x)\big)}{dx} = f'(x) - g'(x).\ _\square\]

Let \(f(x)\) and \(g(x)\) be arbitrary differentiable functions.

The product rule states that

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

Proof:

Recall that for an arbitrary function \(a(x)\),

\[ \frac{d\ \big(a(x)\big)}{dx} = a'(x) = \lim_{h\ \to\ 0} \frac{a(x+h) - a(x)}{h}. \]

Let \(a(x) = f(x) \cdot g(x)\). Substitute \(x+h\) for \(x\) to get \(a(x+h) = f(x+h) \cdot g(x+h)\):

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

Add and subtract \(f(x+h) \cdot g(x)\) in the numerator to get

\[ \begin{align} \frac{d\ \big(f(x) \cdot g(x)\big)}{dx} &= \lim_{h\ \to\ 0} \frac{f(x+h) \cdot g(x) - f(x) \cdot g(x) + f(x+h) \cdot g(x+h) - f(x+h) \cdot g(x)}{h} \\ \\ &= \lim_{h\ \to\ 0} \frac{f(x+h) \cdot g(x) - f(x) \cdot g(x)}{h} + \lim_{h\ \to\ 0} \frac{f(x+h) \cdot g(x+h) - f(x+h) \cdot g(x)}{h} \\ \\ &= \lim_{h\ \to\ 0} \left(g(x) \cdot \frac{f(x+h) - f(x)}{h}\right) + \lim_{h\ \to\ 0} \left(f(x+h) \cdot \frac{g(x+h) - g(x)}{h}\right) \\ \\ &= \left(\lim_{h\ \to\ 0} g(x)\right) \cdot \lim_{h\ \to\ 0} \frac{f(x+h) - f(x)}{h} + \left(\lim_{h\ \to\ 0} f(x+h)\right) \cdot \lim_{h\ \to\ 0} \frac{g(x+h) - g(x)}{h}. \end{align}\]

By definitions of limit and derivative,

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

Let \(f(x)\) and \(g(x)\) be differentiable functions, with \(f(x) \neq 0\) and \(g(x) \neq 0\).

The quotient rule states that

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

Proof:

Recall that for an arbitrary function \(a(x)\),

\[ \frac{d\ \big(a(x)\big)}{dx} = a'(x) = \lim_{h\ \to\ 0} \frac{a(x+h) - a(x)}{h}. \]

Let \(a(x) = \frac{f(x)}{g(x)}\). Substitute \(x+h\) for \(x\) to get \(a(x+h) = \frac{f(x+h)}{g(x+h)}:\)

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

Add and subtract \(f(x) \cdot g(x)\) in the numerator to get

\[ \begin{align} \frac{d\ \left(\frac{f(x)}{g(x)}\right)}{dx} &= \lim_{h\ \to\ 0} \frac{f(x+h) \cdot g(x) - f(x) \cdot g(x) - f(x) \cdot g(x+h) + f(x) \cdot g(x)}{h \cdot g(x+h) \cdot g(x)} \\ &= \lim_{h\ \to\ 0} \frac{\frac{f(x+h)\ \cdot\ g(x) - f(x)\ \cdot\ g(x) - f(x)\ \cdot\ g(x+h) + f(x)\ \cdot\ g(x)}{h}}{g(x+h) \cdot g(x)} \\ \\ &= \lim_{h\ \to\ 0} \frac{\frac{f(x+h)\ \cdot\ g(x) - f(x)\ \cdot\ g(x)}{h} - \frac{f(x)\ \cdot\ g(x+h) - f(x)\ \cdot\ g(x)}{h}}{g(x+h) \cdot g(x)} \\ \\ &= \lim_{h\ \to\ 0} \frac{g(x) \cdot \frac{f(x+h) - f(x)}{h} - f(x) \cdot \frac{g(x+h) - g(x)}{h}}{g(x+h) \cdot g(x)} \\ \\ &= \frac{\displaystyle \lim_{h\ \to\ 0} g(x) \cdot \lim_{h\ \to\ 0} \frac{f(x+h) - f(x)}{h} - \lim_{h\ \to\ 0} f(x) \cdot \lim_{h\ \to\ 0} \frac{g(x+h) - g(x)}{h}}{\displaystyle \lim_{h\ \to\ 0} g(x+h) \cdot \lim_{h\ \to\ 0} g(x)}. \end{align}\]

By definitions of limit and derivative,

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