Calculus

This chapter covers the following topics:

• The Derivative
• L’Hospital’s Rule (L’Hôpital’s Rule)
• Test for a Maximum and Minimum
• Point of Inflection
• Curvature and Radius of Curvature
• Partial Derivative
• Derivatives
• Integral Calculus
• Indefinite Integrals
• Differential Equations
• Fourier Transform
• Laplace Transforms

The derivative

The derivative of a function describes how the function’s output changes as its input changes. In other words, it measures the function’s instantaneous rate of change.

For a function $f(x)$, the derivative is defined by the limit:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

L’Hospital’s rule (L’Hôpital’s rule)

L’Hospital’s Rule helps you evaluate limits that produce indeterminate forms such as $\frac{0}{0}$ or $\frac{\infty }{\infty }$. The idea is to differentiate the numerator and denominator separately.

If

$$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}$$

is indeterminate, then:

$$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

provided the right-hand limit exists.

Test for a maximum and minimum

To decide whether a function $f(x)$ has a local maximum or local minimum at a critical point $x=c$, you can use the second derivative test.

• If $f^{\prime \prime }(c)>0$, then $f(c)$ is a local minimum.
• If $f^{\prime \prime }(c)<0$, then $f(c)$ is a local maximum.

Point of inflection

A point of inflection is a point where the curve changes concavity (the curvature changes direction). In terms of derivatives, this happens when the second derivative changes sign.

A common condition to check is:

$$\frac{d^{2}y}{d{x}^2}=0$$

and there is a sign change in $\frac{d^{2}y}{d{x}^2}$ around that point.

Curvature and radius of curvature

Curvature ($\kappa$) measures how sharply a curve bends at a point. One way to define curvature is:

$$\kappa =\frac{d\theta }{ds}=\frac{|y^{\prime \prime }|}{[1+(y^{\prime }{)}^2{]}^{3/2}}$$

where $\theta$ is the angle of the tangent and $s$ is the arc length.

The radius of curvature ($R$) is the reciprocal of curvature:

$$R=\frac{1}{|\kappa |}$$

It represents the radius of the circular arc that best approximates the curve at that point.

Partial derivative

A partial derivative is the derivative of a multivariable function with respect to one variable, treating the other variables as constants.

For a function $f(x,y)$, the partial derivative with respect to $x$ is:

$$\frac{\partial f}{\partial x}=\frac{d}{dx}f(x,y)$$

Derivatives

In these formulas, $u,v,w$ represent functions of $x$. Also, $a,c$ and $n$ represent constants.

All arguments of the trigonometric functions are in radians. A constant of integration should be added to the integrals.

The following definitions are used:

$$\arcsin u={\sin }^{-1}u,(\sin u{)}^{-1}=\frac{1}{\sin u}$$

Basic differentiation rules

$$\frac{d}{dx}c=0$$

$$\frac{d}{dx}x=1$$

$$\frac{d}{dx}(cu)=c\frac{du}{dx}$$

$$\frac{d}{dx}(u+v-w)=\frac{du}{dx}+\frac{dv}{dx}-\frac{dw}{dx}$$

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

$$\frac{d}{dx}(uvw)=uv\frac{dw}{dx}+uw\frac{dv}{dx}+vw\frac{du}{dx}$$

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

Power and logarithmic differentiation

$$\frac{d}{dx}(u^n)=n{u}^{n-1}\frac{du}{dx}$$

$$\frac{d}{dx}(f(u))=f^{\prime }(u)\frac{du}{dx}$$

$$\frac{du}{dx}=\frac{1}{dx/du}$$

$$\frac{d}{dx}({\log }_{a}u)=({\log }_{a}e)\frac{1}{u}\frac{du}{dx}$$

$$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$

$$\frac{d}{dx}(a^u)=(\ln a)a^u\frac{du}{dx}$$

Exponential and trigonometric differentiation

$$\frac{d}{dx}(e^u)=e^u\frac{du}{dx}$$

$$\frac{d}{dx}(u^u)=u^u\left[\frac{1}{u}\frac{du}{dx}+(\ln u)u^{u-1}\frac{du}{dx}\right]$$

$$\frac{d}{dx}(\sin u)=\cos u\frac{du}{dx}$$

$$\frac{d}{dx}(\cos u)=-\sin u\frac{du}{dx}$$

$$\frac{d}{dx}(\tan u)={\sec }^{2}u\frac{du}{dx}$$

$$\frac{d}{dx}(\cot u)=-{\csc }^{2}u\frac{du}{dx}$$

$$\frac{d}{dx}(\sec u)=\sec u\tan u\frac{du}{dx}$$

$$\frac{d}{dx}(\csc u)=-\csc u\cot u\frac{du}{dx}$$

Inverse trigonometric differentiation

$$\frac{d}{dx}({\sin }^{-1}u)=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},(-\pi /2\le {\sin }^{-1}u\le \pi /2)$$

$$\frac{d}{dx}({\cos }^{-1}u)=\frac{-1}{\sqrt{1-u^2}}\frac{du}{dx},(0\le {\cos }^{-1}u\le \pi )$$

$$\frac{d}{dx}({\tan }^{-1}u)=\frac{1}{1+u^2}\frac{du}{dx},(-\pi /2<{\tan }^{-1}u<\pi /2)$$

$$\frac{d}{dx}({\cot }^{-1}u)=\frac{-1}{1+u^2}\frac{du}{dx},(0<{\cot }^{-1}u<\pi )$$

$$\frac{d}{dx}({\sec }^{-1}u)=\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},(0<{\sec }^{-1}u<\pi /2)\cup (-\pi <{\sec }^{-1}u<-\pi /2)$$

$$\frac{d}{dx}({\csc }^{-1}u)=\frac{-1}{|u|\sqrt{u^2-1}}\frac{du}{dx},(0<{\csc }^{-1}u\le \pi /2)\cup (-\pi <{\csc }^{-1}u\le -\pi /2)$$

Integral calculus

Integral calculus focuses on accumulation. It’s used to compute quantities like area under a curve, volume, and total change.

An indefinite integral represents a family of antiderivatives. For a function $f(x)$:

$$\int f(x)dx=F(x)+C$$

where $F(x)$ is an antiderivative of $f(x)$, and $C$ is the constant of integration.

Indefinite integrals

Basic integration rules

$$\int f^{\prime }(x)dx=f(x)$$

$$\int dx=x$$

$$\int af(x)dx=a\int f(x)dx$$

$$\int [u(x)+v(x)]dx=\int u(x)dx+\int v(x)dx$$

$$\int x^{m}dx=\frac{x^{m+1}}{m+1},(m\ne -1)$$

$$\int u(x)dv(x)=u(x)v(x)-\int v(x)du(x)$$

Logarithmic and exponential integrals

$$\int \frac{dx}{ax+b}=\frac{1}{a}\ln |ax+b|$$

$$\int \frac{dx}{\sqrt{x}}=2\sqrt{x}$$

$$\int a^{x}dx=\frac{a^x}{\ln a}$$

Trigonometric integrals

$$\int \sin xdx=-\cos x$$

$$\int \cos xdx=\sin x$$

$$\int {\sin }^{2}xdx=\frac{x}{2}-\frac{\sin 2x}{4}$$

$$\int {\cos }^{2}xdx=\frac{x}{2}+\frac{\sin 2x}{4}$$

$$\int x\sin xdx=-x\cos x+\sin x$$

$$\int x\cos xdx=\cos x+x\sin x$$

$$\int \sin x\cos xdx=({\sin }^{2}x)/2$$

$$\int \sin ax\cos bxdx=\frac{\cos (a-b)x}{2(a-b)}-\frac{\cos (a+b)x}{2(a+b)},(a^2\ne b^2)$$

Logarithmic and hyperbolic integrals

$$\int \tan xdx=\ln |\cos x|=\ln |\sec x|$$

$$\int \cot xdx=\ln |\sin x|$$

$$\int {\tan }^{2}xdx=\tan x-x$$

$$\int {\cot }^{2}xdx=-\cot x-x$$

$$\int e^{x}dx=(1/a)e^x$$

$$\int e^{ax}dx=\frac{1}{a}{e}^{ax}$$

$$\int \ln xdx=x(\ln x-1)$$

$$\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a},(a>0)$$

Rational integrals

$$\int \frac{dx}{a{x}^2+c}=\frac{1}{\sqrt{ac}}{\tan }^{-1}\left(\sqrt{\frac{a}{c}}x\right),(a>0,c>0)$$

$$\int \frac{dx}{a{x}^2+bx+c}=\frac{2}{\sqrt{4ac-b^2}}{\tan }^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}},(4ac-b^2>0)$$

$$\int \frac{dx}{a{x}^2+bx+c}=\frac{1}{\sqrt{b^2-4ac}}\ln \left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right|,(b^2-4ac>0)$$

$$\int \frac{dx}{a{x}^2+bx+c}=\frac{2}{2ax+b},(b^2-4ac=0)$$

Differential equations

Ordinary linear differential equations

A common class of ordinary linear differential equations has the form:

$$b_n\frac{d^{n}y}{d{x}^n}+b_{n-1}\frac{d^{n-1}y}{d{x}^{n-1}}+\cdots +b_1\frac{dy}{dx}+b_{0}y=f(t)$$

where

$$b_0,b_1,\dots ,b_n$$

are constants.

Homogeneous solution

When the equation is homogeneous (i.e., $f(x)=0$), the solution is:

$$y(x)=C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}+\cdots +C_n{e}^{r_{n}x}$$

where $r_i$ is a distinct root of the characteristic polynomial $P(r)$ with:

$$P(r)=b_n{r}^n+b_{n-1}{r}^{n-1}+\cdots +b_{1}r+b_0$$

Higher orders of multiplicity imply higher powers of $x$. The complete solution for the differential equation is:

$$y(x)=y_h(x)+y_p(x),$$

where $y_h(x)$ is the general solution for the homogeneous equation, and $y_p(x)$ is any particular solution for $f(x)$. If $f(x)$ has $e^{\lambda x}$ terms, then resonance is involved.

Furthermore, specific $f(x)$ forms result in specific $y_p(x)$ forms, some of which are:

• If $f(x)=A$ then $y_p(x)=B$
• If $f(x)=A{e}^{\lambda x}$ then $y_p(x)=B{e}^{\lambda x}$
• If $f(x)=A{x}^m{e}^{\lambda x}$ then $y_p(x)=P_m(x)e^{\lambda x}$, where $P_m(x)$ is a polynomial of degree $m.

If the independent variable is time, $t$, then transient dynamic solutions are implied.

First-order linear homogeneous differential equations with constant coefficients

$$\frac{dy}{dt}+ay=0$$

where $a$ is a real constant.

Solution:

$$y=C{e}^{-at}$$

where $C$ is a constant that satisfies the initial conditions.

First-order linear nonhomogeneous differential equations

$$\frac{dy}{dt}+ay=K{e}^{bt},(a\ne b)$$

with initial condition:

$$y(0)=Kt$$

where, $\tau$ = time constant, and $K$ = gain

Solution:

$$y(t)=Kt+(KB-KC)\left[1-e^{-\frac{t}{\tau }}\right]$$

where:

$$\tau =\frac{1}{a},B=\frac{KB-KC}{a}$$

Second-order linear homogeneous differential equations with constant coefficients

An equation of the form:

$$y^{\prime \prime }+a{y}^{\prime }+by=0$$

can be solved by the method of undetermined coefficients, where a solution of the form:

$$y=C{e}^{rt}$$

is sought.

By substitution, the characteristic equation is:

$$r^2+ar+b=0$$

The roots of the characteristic equation are:

$$r_{1,2}=\frac{-a\pm \sqrt{a^2-4b}}{2}$$

The nature of the solution depends on the discriminant $\Delta =a^2-4b$:

• If $\Delta >0$, the solution is in the form of real distinct roots:

  $$y=C_1{e}^{r_{1}t}+C_2{e}^{r_{2}t}$$

• If $\Delta =0$, the solution is in the form of repeated roots:

  $$y=(C_1+C_{2}t)e^{rt}$$

• If $\Delta <0$, the solution is in the form of complex conjugate roots:

  Given:

  $$r=\alpha \pm i\beta$$

  The solution is:

  $$y=e^{\alpha t}(C_1\cos \beta t+C_2\sin \beta t)$$

where:

$$\alpha =-\frac{a}{2},\beta =\frac{\sqrt{4b-a^2}}{2}$$

Fourier transform

The Fourier transform converts a signal from the time domain to the frequency domain. This is useful because many signals can be viewed as a sum of sinusoids (sines and cosines) at different frequencies. The transform tells you which frequencies are present and how strong they are.

For a continuous-time signal $x(t)$, the Fourier transform is:

$$X(f)={\int }_{-\infty }^{\infty }x(t)\cdot e^{-j2\pi ft}dt$$

Where:

• $X(f)$: Fourier transform of $x(t)$
• $f$: frequency in Hz
• $t$: time in seconds
• $j$: imaginary unit $(\sqrt{-1})$

To retrieve the original time-domain signal $x(t)$ from its frequency-domain representation $X(f)$, use the inverse transform:

$$x(t)={\int }_{-\infty }^{\infty }X(f)\cdot e^{j2\pi ft}df$$

Key properties

Please refer to the FE Handbook for detailed properties of the Fourier Transform.

Laplace transform

The Laplace transform converts a time-domain function (often written as $f(t)$) into a function of a complex variable $s$. It’s widely used in engineering, physics, and control systems, especially for solving differential equations.

The Laplace transform of $f(t)$ is defined as:

$$F(s)={\int }_0^{\infty }f(t)e^{-st}dt$$

and the inverse Laplace transform is:

$$f(t)=\frac{1}{2\pi \mathrm{j}}{\int }_{\sigma -\mathrm{j}\infty }^{\sigma +\mathrm{j}\infty }F(s)e^{st}ds$$

Where:

• $f(t)$: a real- or complex-valued function defined for $t\ge 0$
• $s$: a complex number, $s=\sigma +j\omega$
• $F(s)$: the Laplace transform of $f(t)$

The inverse Laplace transform is used to retrieve the time-domain function $f(t)$ from its Laplace domain $F(s)$.

Often, inverse transforms are found using partial fractions or standard transform tables.

Key properties

Please refer to the FE Handbook for detailed properties of the Laplace transform.