Differential Calculus

Differential calculus is a branch of mathematics that deals with rates of change. The derivative of a function at a chosen input value describes the instantaneous rate of change of the function at that value. The process of finding a derivative is called differentiation. Geometrically, the derivative at a point is the slope of the tangent line to the graph of the function at that point. More generally, for a function \(f\), we can define a derivative function or gradient function, denoted \(f'\), which allows us to calculate the gradient of the tangent at any point on the function.
We have previously differentiated simple functions involving powers of \(x\). In this chapter, we will explore the rules and techniques for differentiating more complicated functions.

The rate of change of a function \(f\) between two points \(A(a, f(a))\) and \(B(b, f(b))\) is the gradient of the secant line \((AB)\).$$ \textcolor{colordef}{\text{Rate of change} = \dfrac{f(b)-f(a)}{b-a}} $$

To find the rate of change at a single point \(A\), we can examine the average rate of change over a very small interval. Let the second point be \(B(a+h, f(a+h))\), where \(h\) is a small change in \(x\). The gradient of the secant line \((AB)\) is given by:$$ \dfrac{f(a+h) - f(a)}{(a+h)-a} = \dfrac{f(a+h) - f(a)}{h} $$As we let \(B\) get closer to \(A\), the value of \(h\) approaches 0. The secant line approaches the tangent line at point A. The limit of the secant slopes is the slope of the tangent, which we define as the instantaneous rate of change.

The derivative of a function \(f\) at a point \(a\), denoted \(f'(a)\), is the instantaneous rate of change of the function at that point. It is defined by the limit:$$ \textcolor{colordef}{f'(a) = \lim_{h \to 0} \dfrac{f(a+h)-f(a)}{h}} $$Geometrically, \(f'(a)\) is the slope of the tangent line to the graph of \(f\) at the point \((a, f(a))\).

Use first principles to find the derivative of \(f(x)=x^2\) at the point \(x=1\).

By finding the derivative at a general point \(x\) instead of a specific point \(a\), we can construct a new function, \(f'(x)\), whose value at any \(x\) is the slope of the tangent to the original function \(f(x)\) at that point.
The process of finding the derivative using this limit is called differentiation from first principles.

The derivative function of \(f\), denoted \(f'\), is the function defined by:$$ f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} $$

For \(f(x)=x^2\), find \(f'(x)\).

While the "prime" notation \(f'(x)\) is compact, an alternative notation developed by Gottfried Wilhelm Leibniz is often more descriptive and versatile, especially when dealing with the chain rule or differential equations.

If \(y\) is a function of \(x\), i.e., \(y=f(x)\), the derivative function can be written as:$$ \textcolor{colordef}{\dfrac{dy}{dx}} $$This is read as "dee y by dee x" and represents the derivative of \(y\) with respect to the variable \(x\).

• The term \(\dfrac{dy}{dx}\) should be thought of as a single operator, not a fraction. However, it is derived from the idea of the fraction \(\dfrac{\Delta y}{\Delta x}\) as the change in \(x\) becomes infinitesimally small.
• We can also use the notation \(\dfrac{d}{dx}[f(x)]\), which is read as "the derivative with respect to \(x\) of \(f(x)\)".

For \(y=x^2\), find \(\dfrac{dy}{dx}\).

Before we explore more advanced differentiation rules, it is important to understand the conditions under which a function can be differentiated. This brings us to the concepts of continuity and differentiability.

• A function is continuous if its graph can be drawn without lifting your pen from the paper. There are no breaks, holes, or jumps.
• A function is differentiable if it is continuous and its graph is "smooth," meaning it has no sharp corners or vertical tangents.

The most important relationship is that differentiability implies continuity. If a function has a well-defined tangent slope at a point, it must be continuous at that point. However, the reverse is not true; a function can be continuous but not differentiable.

A function \(f\) is not differentiable at a point \(x=a\) if its graph has:

• A discontinuity (a hole or jump).
• A sharp corner (where the slope from the left does not equal the slope from the right).
• A vertical tangent (where the slope is infinite).

The graph of a function \(y=f(x)\) is shown. At which x-values is the function not differentiable, and why?

While finding a derivative from first principles using the limit definition is fundamental to understanding the concept, it is often a long and repetitive process. To differentiate more complex functions efficiently, mathematicians have developed a set of powerful rules. This section will introduce these essential rules, which form the bedrock of practical differentiation. By mastering them, you will be able to find the derivative of almost any function you encounter.

We begin with the foundational rules that apply to the most common components of functions, such as constants, powers, and simple arithmetic combinations. These rules can be used to differentiate any polynomial function, as well as many other simple functions, without having to use the limit definition each time.

• Constant Rule: If \(f(x)=c\), then \(f'(x)=0\).
• Power Rule: If \(f(x)=x^n\), then \(f'(x)=nx^{n-1}\) for any \(n \in \mathbb{R}\).
• Constant Multiple Rule: If \(f(x)=c \cdot u(x)\), then \(f'(x)=c \cdot u'(x)\).
• Sum Rule: If \(f(x)=u(x) + v(x)\), then \(f'(x)=u'(x) + v'(x)\).

Find the derivative of \(f(x) = 4x^3 - 5x^2 + 7x - 2\).

Many complex functions are created by composing simpler ones. A composite function has the form \(f(x) = v(u(x))\), where one function (the "inner" function, \(u\)) is the input to another (the "outer" function, \(v\)). For example, in the function \(f(x) = (x^2+1)^3\), the inner function is \(u(x)=x^2+1\) and the outer function is \(v(x)=x^3\).
The Chain Rule provides a powerful method for differentiating such functions by finding the derivatives of the inner and outer functions separately.

If \(f(x)=v(u(x))\) then:$$ f'(x) = v'(u(x)) \cdot u'(x) $$In Leibniz notation, if \(y=v(u)\) then$$ \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} $$

Find the derivative of \(f(x) = (x^2+1)^3\).

While the derivative of a sum is the sum of the derivatives, the same is not true for a product. To find the derivative of a function that is the product of two other functions, such as \(f(x) = x^2 \sin(x)\), we must use a specific formula called the Product Rule.

If \(f(x)=u(x)v(x)\), then$$ f'(x) = u'(x)v(x) + u(x)v'(x) $$In Leibniz notation, if \(y=u\cdot v\) :$$ \dfrac{dy}{dx}= \dfrac{du}{dx}v + u\dfrac{dv}{dx} $$

In words: "The derivative of the first times the second, plus the first times the derivative of the second."

Find the derivative of \(f(x) = (x+1)(x^2+3)\).

Just as with products, finding the derivative of a quotient of two functions requires a specific formula. The Quotient Rule is used to differentiate functions of the form \(f(x) = \dfrac{u(x)}{v(x)}\), such as \(f(x) = \dfrac{e^x}{x^2+1}\).

For \(f(x)=\dfrac{u(x)}{v(x)}\), then$$ f'(x) = \dfrac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$In Leibniz notation, \(y=\dfrac{u}{v}\), then$$ \frac{dy}{dx} = \dfrac{\frac{du}{dx}v - u\frac{dv}{dx}}{v^2} $$

Find the derivative of \(f(x) = \frac{x}{x+1}\).

So far, we have differentiated functions that were defined explicitly, in the form \(y=f(x)\). However, some curves are defined implicitly by a relation between \(x\) and \(y\), such as the circle \(x^2+y^2=25\). In these cases, it may be difficult or impossible to solve for \(y\) directly. Implicit differentiation allows us to find \(\dfrac{dy}{dx}\) without solving for \(y\).

1. Differentiate both sides of the equation with respect to \(x\).
2. When differentiating a term involving \(y\), apply the chain rule. Since \(y\) is a function of \(x\), the derivative of \(y\) is \(\dfrac{dy}{dx}\). For example, \(\dfrac{d}{dx}(y^3) = 3y^2 \cdot \dfrac{dy}{dx}\).
3. After differentiating, solve the resulting equation for \(\dfrac{dy}{dx}\).

Find \(\dfrac{dy}{dx}\) for the circle \(x^2+y^2=3\).

Having established the fundamental rules for combining derivatives (Sum, Product, Quotient, and Chain Rules), we now turn to finding the derivatives of the most important families of functions: exponential, logarithmic, and trigonometric functions.
These functions form the building blocks of most mathematical models in science and engineering. Memorizing their derivatives, along with the rules for combining them, will give you the complete toolkit needed to differentiate nearly any function encountered in this course. Each of these derivatives can be proven from the limit definition, but they are used so frequently that they should be committed to memory.

The exponential function, particularly \(f(x)=e^x\), holds a unique and fundamental place in calculus. Its rate of change is directly proportional to its value, which makes it the mathematical language of many natural processes like population growth and compound interest. We will see that this function has a remarkably simple derivative.

The derivative of the natural exponential function is itself:$$ \dfrac{d}{dx}(e^x) = e^x $$

The derivative of an exponential function with a function in the exponent is given by:$$ \left(e^{u(x)}\right)' = e^{u(x)}u'(x) $$In Leibniz notation:$$ \dfrac{d}{dx}\left(e^{u}\right) = e^{u}\dfrac{du}{dx} $$

Find the derivative of \(f(x)=e^{x^2}\).

For any base \(a>0\), the derivative of \(f(x)=a^x\) is:$$ \dfrac{d}{dx}(a^x) = \ln(a) \cdot a^x $$

The natural logarithm function, \(f(x)=\ln(x)\), is the inverse of the exponential function \(e^x\). This inverse relationship is the key to finding its derivative. Just as the exponential function has a remarkably simple derivative, so does the logarithm. This derivative is fundamental in calculus, particularly in integration, as it allows us to find the antiderivative of functions of the form \(1/x\).

The derivative of the natural logarithm function for \(x>0\) is:$$ \dfrac{d}{dx}(\ln x) = \dfrac{1}{x} $$

The derivative of the natural logarithm of a function is given by:$$ \left(\ln(u(x))\right)' = \dfrac{u'(x)}{u(x)} $$In Leibniz notation:$$ \dfrac{d}{dx}\left(\ln(u)\right) = \dfrac{1}{u}\,\dfrac{du}{dx} $$

Find the derivative of \(f(x) = \ln(2x+3)\).

For complex logarithmic functions, always use the laws of logarithms to expand or simplify the expression before differentiating. This can turn a difficult problem into a very simple one.
Recall the laws: \(\ln(ab)=\ln(a)+\ln(b)\), \(\ln(a/b)=\ln(a)-\ln(b)\), and \(\ln(a^n)=n\ln(a)\).

Find the derivative of \(f(x) = \ln\left(\dfrac{x+1}{x}\right)\).

For any base \(a>0, a\neq 1\), the derivative of \(f(x)=\log_a(x)\) is:$$ \dfrac{d}{dx}(\log_a x) = \dfrac{1}{x\ln(a)} $$

Trigonometric functions describe periodic phenomena, which are common in physics and engineering. Their derivatives follow a cyclical pattern that is both elegant and powerful. The proofs for the derivatives of \(\sin(x)\) and \(\cos(x)\) are derived from the limit definition and rely on two fundamental trigonometric limits.

For \(x\) in radians:$$ \lim_{x \to 0} \dfrac{\sin(x)}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \dfrac{\cos(x)-1}{x} = 0 $$

The derivatives of the basic trigonometric functions are:$$ \dfrac{d}{dx}(\sin x) = \cos x \quad \text{and} \quad \dfrac{d}{dx}(\cos x) = -\sin x $$

The derivative of the tangent function is:$$ \dfrac{d}{dx}(\tan x) = 1+\tan^2 x $$

• \((\sin(u(x)))' = u'(x)\cos(u(x))\)
• \((\cos(u(x)))' = -u'(x)\sin(u(x))\)
• \((\tan(u(x)))' = u'(x)\bigl(1+\tan^2(u(x))\bigr)\)

In Leibniz notation:

• \(\dfrac{d}{dx}(\sin(u)) = \cos(u)\dfrac{du}{dx}\)
• \(\dfrac{d}{dx}(\cos(u)) = -\sin(u)\dfrac{du}{dx}\)
• \(\dfrac{d}{dx}(\tan(u)) = \bigl(1+\tan^2(u)\bigr)\dfrac{du}{dx}\)

Find the derivative of \(f(x) = \cos(x^2+3x)\).

The derivatives of the remaining trigonometric functions—secant, cosecant, and cotangent—can be found by expressing them in terms of sine and cosine and then applying the Quotient Rule.

$$ \dfrac{d}{dx}(\sec x) = \sec x \tan x $$$$ \dfrac{d}{dx}(\csc x) = -\csc x \cot x $$$$ \dfrac{d}{dx}(\cot x) = -\csc^2 x $$

The inverse trigonometric functions (\(\arcsin\), \(\arccos\), \(\arctan\)) are essential in problems involving angles. Their derivatives are algebraic functions and are found using implicit differentiation.

$$ \dfrac{d}{dx}(\arcsin x) = \dfrac{1}{\sqrt{1-x^2}} $$$$ \dfrac{d}{dx}(\arccos x) = -\dfrac{1}{\sqrt{1-x^2}} $$$$ \dfrac{d}{dx}(\arctan x) = \dfrac{1}{1+x^2} $$

Find the derivative of \(f(x) = \arctan(e^x)\).

The second derivative of \(f\), denoted \(f''\), is the derivative of the first derivative, \(f'\).$$ f''(x) = \dfrac{d}{dx}(f'(x)) \quad \text{ or in Leibniz notation, } \quad \dfrac{d^2f}{dx^2}= \dfrac{d}{dx}\left(\dfrac{df}{dx} \right) $$

Find the second derivative of \(f(x)=x^4 - 5x^2\).

When evaluating limits, direct substitution is the easiest method. However, this method sometimes results in an indeterminate form, such as \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\). These forms do not have a defined value, but the limit might still exist.
For example, \(\displaystyle\lim_{x \to 0} \dfrac{e^x-1}{x}\) gives \(\dfrac{0}{0}\). L'Hôpital's Rule provides a powerful technique to solve limits of these specific indeterminate forms by using derivatives.

Suppose \(f\) and \(g\) are differentiable functions and \(g'(x)\neq 0\) on an open interval \(I\) containing \(a\) (except possibly at \(a\) itself).

• Indeterminate Form \(\dfrac{0}{0}\):
  If \(\displaystyle\lim_{x \to a} f(x) = 0\), \(\displaystyle\lim_{x \to a} g(x) = 0\) and \(\lim_{x \to a} \dfrac{f'(x)}{g'(x)}\) exists, then $$ \lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)} $$
• Indeterminate Form \(\dfrac{\infty}{\infty}\):
  If \(\displaystyle\lim_{x \to a} f(x) = \pm\infty\), \(\displaystyle\lim_{x \to a} g(x) = \pm\infty\) and \(\lim_{x \to a} \dfrac{f'(x)}{g'(x)}\) exists, then $$ \lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)} $$

Evaluate the limit \(\displaystyle\lim_{x \to 0} \dfrac{e^{x}-1}{x}\).