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Calculus Review

Table of Contents

Differentiation

Definition

Let ff be a function, and define the limit

f′(x)=lim⁡h→0f(x+h)−f(x)h.f'\p{x} = \lim_{h\to0} \frac{f\p{x + h} - f\p{x}}{h}.

If this limit exists, then we say that ff is differentiable at xx and we call f′f' the derivative of ff. f′f' is also written as ddxf(x)\deriv{}{x} f\p{x} or dfdx\deriv{f}{x}.
Proposition (properties of the derivative)

Let f,gf, g be differentiable functions and cc be a real number. Then the following hold:

  1. ddx\deriv{}{x} is linear, that is,

    ddx(f(x)+g(x))=ddxf(x)+ddxg(x)andddxcf(x)=cddxf(x).\deriv{}{x} \p{f\p{x} + g(x)} = \deriv{}{x} f\p{x} + \deriv{}{x} g\p{x} \quad\text{and}\quad \deriv{}{x} cf\p{x} = c \deriv{}{x} f\p{x}.

  2. The product rule:

    ddxf(x)g(x)=f′(x)g(x)+f(x)g′(x)\deriv{}{x} f\p{x}g\p{x} = f'\p{x}g\p{x} + f\p{x}g'\p{x}

  3. The chain rule:

    ddxf(g(x))=f′(g(x))g′(x)\deriv{}{x} f\p{g\p{x}} = f'\p{g\p{x}}g'\p{x}

Another commonly used property is the quotient rule:

ddxf(x)g(x)=f′(x)g(x)−g′(x)f(x)[g(x)]2,\deriv{}{x} \frac{f\p{x}}{g\p{x}} = \frac{f'\p{x}g\p{x} - g'\p{x}f\p{x}}{\br{g\p{x}}^2},

when g(x)g\p{x} is non-zero.

Exercise 1.

By writing f(x)g(x)=f(x)[g(x)]−1\frac{f\p{x}}{g\p{x}} = f\p{x}\br{g\p{x}}^{-1}, derive the quotient rule using the product rule and the chain rule.
Proposition

Common Derivatives

ddxxn=−nxn−1ddxsin⁡x=−cos⁡xddxcos⁡x=−sin⁡xddxtan⁡x=−sec⁡2xddxcot⁡x=−csc⁡2xddxsec⁡x=−sec⁡xtan⁡xddxcsc⁡x=−csc⁡xcot⁡xddxex=−exddxlog⁡x=−1x\begin{aligned} &\deriv{}{x} x^n &=& \phantom{-}nx^{n-1} \\[2ex] &\deriv{}{x} \sin{x} &=& \phantom{-}\cos{x} \\[2ex] &\deriv{}{x} \cos{x} &=& -\sin{x} \\[2ex] &\deriv{}{x} \tan{x} &=& \phantom{-}\sec^2{x} \\[2ex] &\deriv{}{x} \cot{x} &=& -\csc^2{x} \\[2ex] &\deriv{}{x} \sec{x} &=& \phantom{-}\sec{x}\tan{x} \\[2ex] &\deriv{}{x} \csc{x} &=& -\csc{x}\cot{x} \\[2ex] &\deriv{}{x} e^x &=& \phantom{-}e^x \\[2ex] &\deriv{}{x} \log{x} &=& \phantom{-}\frac{1}{x} \end{aligned}

Note that log⁡x\log{x} is base ee, not\textbf{not} base 1010.
Example 1.

Calculate ddxlog⁡ax\deriv{}{x} \log_a{x}, where a>0a > 0 and a≠1a \neq 1.
Solution.

By the change of base formula, we have log⁡ax=log⁡xlog⁡a\log_a{x} = \frac{\log{x}}{\log{a}}. Referring to the table above, we get

ddxlog⁡ax=ddxlog⁡xlog⁡a=1xlog⁡a.\deriv{}{x} \log_a{x} = \deriv{}{x} \frac{\log{x}}{\log{a}} = \boxed{\frac{1}{x\log{a}}}.
Exercise 2.

Calculate ddxbx\deriv{}{x} b^x, where b>0b > 0. (Hint: b=elog⁡bb = e^{\log{b}})

Integration

Definition

Let ff be continuous. If FF is a function which satisfies F′=fF' = f, then FF is called an antiderivative or integral of ff. Furthermore, indefinite integral of ff with respect to xx is

∫f(x) dx=F(x)+C,\int f\p{x} \,\diff{x} = F\p{x} + C,

where CC is a constant.
Proposition (properties of the integral)

Let f,gf, g be integrable functions and cc be a real number. Then the following hold:

  1. Integration is linear, that is,

    ∫f(x)+g(x) dx=∫f(x) dx+∫g(x) dxand∫cf(x) dx=c∫f(x) dx.\int f\p{x} + g\p{x} \,\diff{x} = \int f\p{x} \,\diff{x} + \int g\p{x} \,\diff{x} \quad\text{and}\quad \int cf\p{x} \,\diff{x} = c \int f\p{x} \,\diff{x}.

  2. ∫abf(x) dx=−∫baf(x) dx\int_a^b f\p{x} \,\diff{x} = -\int_b^a f\p{x} \,\diff{x}

  3. ∫abf(x) dx+∫bcf(x) dx=∫acf(x) dx\int_a^b f\p{x} \,\diff{x} + \int_b^c f\p{x} \,\diff{x} = \int_a^c f\p{x} \,\diff{x}
Theorem (the fundamental theorem of calculus)

Let FF be an antiderivative of ff. Then

  1. ∫abf(x) dx=F(b)−F(a)\int_a^b f\p{x} \,\diff{x} = F\p{b} - F\p{a}
  2. ddx∫axf(t) dt=f(x)\deriv{}{x} \int_a^x f\p{t} \,\diff{t} = f\p{x}

Notice that in (ii), we replaced xx with tt as the variable of integration because the bounds contain xx. Depending on who you ask, using xx as a bound and the variable of integration at the same time may be considered incorrect (i.e., if you do this, you might lose points on exams!). Either way, it can become a big source of confusion, so it's good to get in the habit of doing this.

The variable of integration (xx in (i) and tt in (ii)) is called a dummy variable, which just means that the symbol you put in their place doesn't matter as long as it isn't used in the bounds. For example, it's perfectly valid to write

ddx∫axf(😊) d😊=f(x),\deriv{}{x} \int_a^x f\p{😊} \,\diff{😊} = f\p{x},

and you should avoid writing anything like

∫a😢f(😢) d😢.\int_a^😢 f\p{😢} \,\diff{😢}.

Integration can be thought of as the inverse of differentiation. For example,

ddxxn=nxn−1  ⟹  ∫nxn−1 dx=xn+C.\deriv{}{x} x^n = nx^{n-1} \implies \int nx^{n-1} \,\diff{x} = x^n + C.
Exercise 3.

Transform the derivative table into an integration table.
Example 2.

Calculate ∫tan⁡x dx\int \tan{x} \,\diff{x}.
Solution.

Recall that tan⁡x=sin⁡xcos⁡x\tan{x} = \frac{\sin{x}}{\cos{x}}. We'll use uu-substitution: set u=cos⁡xu = \cos{x}, which gives du=−sin⁡x dx\diff{u} = -\sin{x} \,\diff{x}. Then

∫tan⁡x dx=∫sin⁡xcos⁡x dx=∫1cos⁡xsin⁡(x) dx=−∫1cos⁡x−sin⁡(x) dx=−∫1u du=−log⁡∣u∣+C=−log⁡∣cos⁡x∣+C.\begin{aligned} \int \tan{x} \,\diff{x} &= \int \frac{\sin{x}}{\cos{x}} \,\diff{x} \\[2ex] &= \int \frac{1}{\cos{x}} \sin\p{x} \,\diff{x} \\[2ex] &= -\int \frac{1}{\colorbox{red}{$\cos{x}$}} \colorbox{yellow}{$-\sin\p{x} \,\diff{x}$} \\[2ex] &= -\int \frac{1}{\colorbox{red}{$u$}} \colorbox{yellow}{$\,\diff{u}$} \\[2ex] &= -\log{\abs{u}} + C \\[2ex] &= \boxed{-\log{\abs{\cos{x}}} + C}. \end{aligned}
Exercise 4.

Calculate ∫cot⁡x dx\int \cot{x} \,\diff{x}.