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
• Differentiation rules
• Integral calculus
• Indefinite integrals
• Differential equations

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.

Example: evaluating a limit with L’Hospital’s rule

Evaluate ${\lim }_{x\to 0}\frac{\sin x}{x}$.

Direct substitution gives $\frac{0}{0}$, an indeterminate form, so L’Hospital’s rule applies. Differentiate the numerator and denominator separately, then substitute again:

$$\underset{x\to 0}{\lim }\frac{\sin x}{x}=\underset{x\to 0}{\lim }\frac{\cos x}{1}=\cos (0)=1$$

Answer: $1$

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.

A critical point is where $f^{\prime }(x)=0$ (or where $f^{\prime }(x)$ does not exist), and the test is inconclusive when $f^{\prime \prime }(c)=0$. On a closed interval $[a,b]$, the absolute maximum or minimum can also fall at an endpoint, so evaluate $f$ at every critical point inside the interval and at both endpoints, then compare.

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:

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

But this condition alone isn’t sufficient - you also need to confirm that $\frac{d^{2}y}{d{x}^2}$ actually changes sign around that point. If it doesn’t, the point isn’t a true inflection point even though the second derivative is zero there.

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.

Example: radius of curvature

Find the radius of curvature of $y=x^2$ at $x=1$.

First find the derivatives: $y^{\prime }=2x$, so $y^{\prime }(1)=2$, and $y^{\prime \prime }=2$.

Substitute into the curvature formula:

$$\kappa =\frac{|2|}{[1+(2{)}^2{]}^{3/2}}=\frac{2}{5^{3/2}}\approx \frac{2}{11.18}\approx 0.179$$

The radius of curvature is the reciprocal:

$$R=\frac{1}{0.179}\approx 5.59$$

Answer: $R\approx 5.59$

Derivatives

In these formulas, $u,v,w$ represent functions of $x$. Also, $a,c$, and $n$ represent constants. These tables mirror the FE Reference Handbook’s Mathematics section, so the exam skill to build is locating and applying the right entry quickly, not memorizing the whole list.

Exam tip: All trigonometric arguments are in radians, not degrees, and remember to add a constant of integration, $C$, to indefinite integrals. Also carry full calculator precision through each intermediate step rather than rounding early - small rounding errors compound quickly across multi-step derivative and integral problems.

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

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

$$\frac{d}{dx}(u^v)=v{u}^{v-1}\frac{du}{dx}+u^v(\ln u)\frac{dv}{dx}$$

With $v=u$, this gives $\frac{d}{dx}(u^u)=u^u(1+\ln u)\frac{du}{dx}$.

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

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

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

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

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

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

Inverse trigonometric differentiation

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

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

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

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

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

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

Example: logarithmic differentiation of $y=(2x{)}^x$

Take the natural log of both sides: $\ln y=x\ln (2x)$.

Differentiate both sides with respect to $x$, applying the product rule on the right side:

$$\frac{1}{y}\frac{dy}{dx}=\ln (2x)+x\cdot \frac{1}{x}=\ln (2x)+1$$

Multiply both sides by $y=(2x{)}^x$ to isolate $\frac{dy}{dx}$:

$$\frac{dy}{dx}=(2x{)}^x\left[\ln (2x)+1\right]$$

Answer: $\frac{dy}{dx}=(2x{)}^x[\ln (2x)+1]$

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.

Example: area between two curves

Find the area of the region bounded by $y=x$ and $y=x^2$ between their intersection points.

The curves intersect where $x=x^2$, so $x=0$ and $x=1$. On $[0,1]$, $y=x$ lies above $y=x^2$, so the area between them is the definite integral of their difference:

$$A={\int }_0^1(x-x^2)dx={\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$$

Answer: $\frac{1}{6}$

Indefinite integrals

Basic integration rules

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

$$\int dx=x$$

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

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

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

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

Example: integration by parts

Find $\int x\sin xdx$.

Choose $u=x$, which gets simpler when differentiated, and $dv=\sin xdx$, so $du=dx$ and $v=-\cos x$. Rule 6 gives:

$$\int x\sin xdx=-x\cos x-\int (-\cos x)dx=-x\cos x+\sin x$$

Answer: $-x\cos x+\sin x$ (plus a constant), which is entry 14 below. Choosing $u=\sin x$ and $dv=xdx$ instead leaves $\int \frac{x^2}{2}\cos xdx$, a harder integral than the one you started with - the sign that $u$ and $dv$ are the wrong way round.

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

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

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

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

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

$$\int e^{x}dx=e^x$$

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

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

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

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:

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

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 $f(x)$.

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 $m$.

First-order linear homogeneous differential equations with constant coefficients

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

where $a$ is a real constant. The solution is:

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

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

First-order linear nonhomogeneous differential equations

$$\tau \frac{dy}{dt}+y=Kx(t)$$

where $\tau$ is the time constant, $K$ is the gain, and $x(t)=A$ is a constant forcing input. With initial condition $y(0)=y_0$, the solution is:

$$y(t)=KA+(y_0-KA)e^{-t/\tau }$$

Dividing through by $\tau$ puts the equation in the form $\frac{dy}{dt}+ay=aKx(t)$, so $\tau =\frac{1}{a}$ is the time constant.

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 assuming a trial exponential solution:

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

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:

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

Example: solving a second-order homogeneous ODE

Solve $y^{\prime \prime }-3y^{\prime }+2y=0$.

The characteristic equation is $r^2-3r+2=0$, which factors as $(r-1)(r-2)=0$, giving roots $r_1=1$ and $r_2=2$. Since the roots are real and distinct, the general solution is:

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

Answer: $y=C_1{e}^t+C_2{e}^{2t}$