Differentiation Techniques: Notation, Rules, and Special Functions

Notation & Definition of the Derivative

Understanding the Derivative

• The operator $\frac{d}{dx}$ means “differentiate (something) with respect to $x$.” It is meaningless on its own.

• Given a function $y(x)$, the derivative of this function can be written in three ways:

  • $\frac{dy}{dx}$

  • $y'(x)$

  • $\frac{d}{dx} y(x)$

• $\frac{dy}{dx}$ is the slope of a tangent line to the graph of $y$ at a particular $x$, or the instantaneous rate of change of $y$ with respect to $x$.

Standard Derivatives

Common Derivative Formulas

• $\frac{d}{dx}(x^n) = n x^{n-1}$, $n \neq 0$

• $\frac{d}{dx}(e^x) = e^x$

• $\frac{d}{dx}(\sin x) = \cos x$

• $\frac{d}{dx}(\cos x) = -\sin x$

• $\frac{d}{dx}(\tan x) = \sec^2 x$

Derivatives of Sums and Scalar Multiples

Linearity of the Derivative

• $\frac{d}{dx}(c f(x)) = c \frac{d}{dx}(f(x))$, where $c \in \mathbb{R}$

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

Product Rule

Derivative of a Product

• If $u(x)$ and $v(x)$ are differentiable functions, then:

$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

Quotient Rule

Derivative of a Quotient

• If $u(x)$ and $v(x)$ are differentiable functions, then:

$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Chain Rule

Derivative of a Composite Function

• If $y$ can be written as a function of $u$ (the ‘inner function’), where $u$ is a function of $x$, then:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

• In general, we do not need to use labels such as $u$. Instead, we think in terms of ‘inner’ and ‘outer’ functions.

• Hence, $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$

The derivative of a composite function is equal to the derivative of the outer function (evaluated at the inner function) times the derivative of the inner function.

Implicit Differentiation

Differentiating Implicit Relations

• Used to find derivatives of general relations of the form $F(x, y) = 0$ that cannot be rearranged into the form $y = f(x)$.

• We say that $F(x, y) = 0$ implicitly defines $y$ to be a function of $x$.

Procedure: Given the implicit equation $F(x, y) = 0$:

1. Differentiated both sides of the equation with respect to $x$ (without thinking of $y$ as a function of $x$). Expressions involving $y$ require the use of the chain rule.

2. Collect the terms in $dy/dx$ on one side of the equation.

3. Factor out $dy/dx$.

4. Divide both sides by the coefficient of $dy/dx$.

Logarithmic Differentiation

Using Logarithms to Simplify Differentiation

• Take the logarithm of a complex expression (involving products, quotients, and exponents) to differentiate it more easily (thereby splitting it into more manageable parts) and then differentiate implicitly.

• The general power function, $y = [g(x)]^{h(x)}$, cannot be differentiated using the Power Rule for derivatives. Instead, use logarithmic differentiation.

• First take natural logs of both sides: $\ln(y) = g(x) \cdot \ln(f(x))$, then differentiate both sides.

Derivatives of Hyperbolic Functions

Standard Hyperbolic Derivatives

Function	Derivative
$\sinh x$	$\cosh x$
$\cosh x$	$\sinh x$
$\tanh x$	$\text{sech}^2 x$
$\text{sech} x$	$-\text{sech} x \tanh x$
$\text{csch} x$	$-\text{csch} x \coth x$
$\coth x$	$-\text{csch}^2 x$

Derivatives of Inverse Hyperbolic Functions

Standard Inverse Hyperbolic Derivatives

Function	Derivative
$\sinh^{-1} x$	$\frac{1}{\sqrt{x^2 + 1}}$
$\cosh^{-1} x$	$\frac{1}{\sqrt{x^2 - 1}}$
$\tanh^{-1} x$	$\frac{1}{1 - x^2}$
$\text{sech}^{-1} x$	$\frac{-1}{x \sqrt{1 - x^2}}$
$\text{csch}^{-1} x$	$\frac{-1}{|x| \sqrt{1 + x^2}}$
$\coth^{-1} x$	$\frac{1}{1 - x^2}$

Note: All derivatives are given on the 6-page MATH1020 Calculus for Engineers Formulae sheet.