Techniques of Differentiation

Introduction to differentiation techniques

In calculus, differentiation is used to find the rate of change of a function. There are several techniques to differentiate different types of functions. These techniques simplify the process of finding derivatives and are especially useful for complex expressions. Common techniques include the power rule, product rule, quotient rule, chain rule, and implicit differentiation.

Power rule

The power rule is one of the simplest techniques for differentiating polynomial functions. The rule states that for a function of the form:

$f(x) = ax^n$

The derivative of $f(x)$ with respect to $x$ is:

$f'(x) = n \cdot ax^{n-1}$

In other words, to differentiate $ax^n$, multiply the exponent $n$ by the coefficient $a$, then reduce the exponent by 1.

Example 1: Differentiate $f(x) = 5x^3$

We will apply the power rule: Identify the coefficient and exponent: $a = 5$, $n = 3$. Use the power rule formula: $f'(x) = n \cdot ax^{n-1}$.

$f'(x) = 3 \cdot 5x^{3-1}$

Simplify the expression:

$f'(x) = 15x^2$

Example 2: Differentiate $f(x) = -2x^4$

Apply the power rule: Identify the coefficient and exponent: $a = -2$, $n = 4$. Use the power rule formula:

$f'(x) = 4 \cdot (-2)x^{4-1}$

Simplify the expression:

$f'(x) = -8x^3$

Product rule

The product rule is used when differentiating the product of two functions. If $f(x) = g(x) \cdot h(x)$, then the product rule states:

$f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)$

That is, to differentiate the product of two functions, differentiate the first function, multiply by the second function, and add the first function multiplied by the derivative of the second function.

Example 1: Differentiate $f(x) = x^2 \cdot \sin(x)$

We will apply the product rule:

1. Let $g(x) = x^2$ and $h(x) = \sin(x)$.
2. Differentiate $g(x)$: $g'(x) = 2x$.
3. Differentiate $h(x)$: $h'(x) = \cos(x)$.
4. Now apply the product rule:

$f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)$

$f'(x) = 2x \cdot \sin(x) + x^2 \cdot \cos(x)$

Example 2: Differentiate $f(x) = e^x \cdot \ln(x)$

Apply the product rule:

1. Let $g(x) = e^x$ and $h(x) = \ln(x)$.
2. Differentiate $g(x)$: $g'(x) = e^x$.
3. Differentiate $h(x)$: $h'(x) = 1/x$.
4. Now apply the product rule:

$f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)$

$f'(x) = e^x \cdot \ln(x) + e^x \cdot (1/x)$

Quotient rule

The quotient rule is used when differentiating a function that is the quotient of two other functions. If $f(x) = g(x) / h(x)$, then the quotient rule states:

$f'(x) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{h(x)^2}$

That is, to differentiate a quotient, differentiate the numerator and denominator, multiply them accordingly, and divide by the square of the denominator.

Example 1: Differentiate $f(x) = (x^2 + 1) / (x - 1)$

Apply the quotient rule:

1. Let $g(x) = x^2 + 1$ and $h(x) = x - 1$.
2. Differentiate $g(x)$: $g'(x) = 2x$.
3. Differentiate $h(x)$: $h'(x) = 1$.
4. Now apply the quotient rule:

$f'(x) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{h(x)^2}$

$f'(x) = \frac{2x \cdot (x - 1) - (x^2 + 1) \cdot 1}{(x - 1)^2}$

Simplify the expression:

$f'(x) = \frac{2x^2 - 2x - x^2 - 1}{(x - 1)^2}$

$f'(x) = \frac{x^2 - 2x - 1}{(x - 1)^2}$

Chain rule

The chain rule is used to differentiate composite functions. If $f(x) = g(h(x))$, then the chain rule states:

$f'(x) = g'(h(x)) \cdot h'(x)$

That is, differentiate the outer function $g$ with respect to $h(x)$, then multiply by the derivative of the inner function $h(x)$.

Example 1: Differentiate $f(x) = \sin(3x)$

Apply the chain rule:

1. Let $g(u) = \sin(u)$ and $h(x) = 3x$.
2. Differentiate $g(u)$: $g'(u) = \cos(u)$.
3. Differentiate $h(x)$: $h'(x) = 3$.
4. Now apply the chain rule:

$f'(x) = g'(h(x)) \cdot h'(x)$

$f'(x) = \cos(3x) \cdot 3$

$f'(x) = 3\cos(3x)$

Example 2: Differentiate $f(x) = e^{x^2}$

Apply the chain rule:

1. Let $g(u) = e^u$ and $h(x) = x^2$.
2. Differentiate $g(u)$: $g'(u) = e^u$.
3. Differentiate $h(x)$: $h'(x) = 2x$.
4. Now apply the chain rule:

$f'(x) = g'(h(x)) \cdot h'(x)$

$f'(x) = e^{x^2} \cdot 2x$

$f'(x) = 2x \cdot e^{x^2}$

Implicit differentiation

Implicit differentiation is used when the function is not given in the form $y = f(x)$, but rather in a form where both $x$ and $y$ are mixed together. The goal is to differentiate with respect to $x$ while treating $y$ as a function of $x$.

Example 1: Differentiate $x^2 + y^2 = 25$

Apply implicit differentiation: Differentiate both sides with respect to $x$:

$2x + 2y \cdot \frac{dy}{dx} = 0$

Isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-2x}{2y}$

$\frac{dy}{dx} = \frac{-x}{y}$