Derivatives

MATH1051

Derivatives

Finding the instantaneous velocity of a moving object and other problems involving rates of change are situations where derivatives can be used as a powerful tool. All rates of change can be interpreted as slopes of appropriate tangents. Therefore we shall consider the tangent problem and how it leads to a precise definition of the derivative.

6.1 Tangents

6.2 Definition of Derivative

The derivative of $f$ at $x$ is defined by

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

We say that $f$ is differentiable at some point $x$ if this limit exists. Further, we say that $f$ is differentiable on an open interval if it is differentiable at every point in the interval. Note that $f^{\prime }(a)$ is the slope of the tangent line to the graph of $y=f(x)$ at $x=a.$

We have thus defined a new function $f^{\prime }$, called the derivative of $f$. Sometimes we use the Leibniz notation $\frac{dy}{dx}$ or $\frac{df}{dx}$ in place of $f^{\prime }(x).$

Note that if $f$ is differentiable at $a$, there holds:

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

6.2.1 Example

Using the definition of the derivative (“from first principles”), find the derivative of $f(x)=x^2+x$.

Using the definition for a derivative, we find:

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{(x+h{)}^2+(x+h)-(x^2+x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{x^2+2xh+h^2+x+h-x^2-x}{h}\\ & =\underset{h\to 0}{\lim }\frac{2xh+h^2+h}{h}\\ & =\underset{h\to 0}{\lim }(2x+h+1)\\ & =2x+1\end{array}$$

6.4 Chain Rule

8.2 The Derivative of a Function

The derivative of a function $f$ at the number $a$, denoted by $f^{\prime }(a)$ is

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

if this limit exists.

Thus the tangent line to the curve $y=f(x)$ at $x=a$ is the line through $(a,f(a))$ whose slope is equal to $f^{\prime }(a)$, the derivative of $f$ at $a$.

The derivative $f^{\prime }(a)$ is the instantaneous rate of change of $y=f(x)$ with respect to $x$ when $x=a$.

Given a function $f$, we can define a new function, called the derivative of $f$, and denoted $f^{\prime }$, by the rule that

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

The domain of $f^{\prime }$ is the set of all $x$ in the domain of $f$ such that $f^{\prime }(x)$ exists.

The derivative of $f$ is also called the derived function of $f$. The process of determining the derivative of a function $f$ is called differentiation.

If $y=f(x)$, then other notation commonly used for the derivative $f^{\prime }(x)$ includes:

$$y^{\prime }\frac{dy}{dx}\frac{df}{dx}\frac{d}{dx}f(x)$$

A function $f$ is said to be differentiable at $a$ if $f^{\prime }(a)$ exists.

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8.3 Differentiation Rules

Determining derivatives from the definition can be time-consuming. Luckily there are some rules for differentiation that speed up the process. The proofs of these rules can be found in most calculus textbooks, but you do not need to know the proofs for this course.

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The derivative of a constant function is zero.

If $f(x)=x^n$, then $f^{\prime }(x)=n{x}^{n-1}$ for $n\in \mathbb{R}$.

Constant Multiple Rule

If $c$ is a constant and $f$ is a differentiable function, then

$$\frac{d}{dx}(cf(x))=c\frac{d}{dx}f(x)$$

Sum Rule

If $f$ and $g$ are both differentiable functions, then

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

This rule can also be written as $(f+g{)}^{\prime }=f^{\prime }+g^{\prime }$.

Product Rule

If $f$ and $g$ are both differentiable functions, then

$$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{\left(g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)\right)}{(g(x){)}^2}$$

Chain Rule (Composite Function Rule)

If $f$ and $g$ are both differentiable functions, then

$$(f\circ g{)}^{\prime }(x)=f^{\prime }(g(x))g^{\prime }(x)$$

If $y=f(u)$ and $u=g(x)$, then this rule can be written as

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

To apply the chain rule, think about starting with the outside function and working your way in.

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So far we have only looked at deriving relations where $y$ or $f(x)$ is defined explicitly as a relation of $x$. What happens if $y$ is defined implicitly as a relation of $x$, such as $y^2=x$ or $y^3+2y=4x^3$? We can use a technique called implicit differentiation which is based on the chain rule.

Let’s look at the derivative of $x^2+y^2=25$ (Circle, radius 5, centred at the origin). We differentiate both sides with respect to $x$,

$$\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}25$$ $$⟹\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=0$$

We can easily work out $\frac{d}{dx}(x^2)$, but how can we do $\frac{d}{dx}(y^2)$?

This is where the chain rule comes in.

$$\frac{d}{dx}(y^2)=\frac{d}{dy}(y^2)\frac{dy}{dx}=2y\frac{dy}{dx}$$

So we now have $2x+2y\frac{dy}{dc}=0$

$$⟹\frac{dy}{dx}=-\frac{2x}{2y}$$ $$⟹\frac{dy}{dx}=-\frac{x}{y}$$

So the derivative of $y$ with respect to $x$, where $y$ is defined in terms of $x$ implicitly by the equation $x^2+y^2=25$, is $\frac{dy}{dx}=-\frac{x}{y}$.

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Derivatives of Trigonometric Functions

Note that when we use the trigonometric functions, such as $f(x)=\sin x$, all angles are measured in radians.

To determine the derivative of $f(x)=\sin x$ we require two special limits:

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

The proofs of these limits can be found in most calculus textbooks, but we won’t worry about the proofs in this course.

If $f(x)=\sin x$ then $f^{\prime }(x)=\cos x$.

Proof

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{\sin (x+h)-\sin x}{h}\\ & =\underset{h\to 0}{\lim }\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ & =\underset{h\to 0}{\lim }\left(\frac{\sin x\cos h-\sin x}{h}+\frac{\cos x\sin h}{h}\right)\\ & =\underset{h\to 0}{\lim }\left(\sin x\left(\frac{\cos h-1}{h}\right)+\cos x\left(\frac{\sin h}{h}\right)\right)\\ & =\underset{h\to 0}{\lim }\sin x\underset{h\to 0}{\lim }\left(\frac{\cos h-1}{h}\right)+\underset{h\to 0}{\lim }\cos x\underset{h\to 0}{\lim }\left(\frac{\sin h}{h}\right)\\ & =\sin x\cdot 0+\cos x\cdot 1\\ & =\cos x\end{array}$$

Using a similar method to the previous page, we can show that if $f(x)=\cos x$ then $f^{\prime }(x)=-\sin x$.

The derivatives of the trigonometric functions are summarised below:

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$$\frac{d}{dx}(\sin x)=\cos x$$ $$\frac{d}{dx}(\cos x)=-\sin x$$ $$\frac{d}{dx}(\tan x)=se{c}^{2}x$$ $$\frac{d}{dx}(\csc x)=-\csc x\cot x$$ $$\frac{d}{dx}(\sec x)=\sec x\tan x$$ $$\frac{d}{dx}(\cot x)=-{\csc }^{2}x$$

If $f(x)=e^x$, then $f^{\prime }(x)=e^x$. Note: Another way of writing $e^x$ is $\exp (x)$.

Thus, on the graph of $f(x)=e^x$, the slope of the tangent line at each point is equal to the function value at that point.

If $f(x)=\ln x$, then $f^{\prime }(x)=\frac{1}{x}$. Note that here $x>0$ since the domain of $f(x)=\ln x$ is $x>0$.

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6.5 Derivative of an Inverse Function

Suppose $y=f^{-1}(x)$, where $f^{-1}$ is the inverse of $f$. To obtain $\frac{dy}{dx}$ we use

$$x=f(f^{-1}(x))=f(y)$$

Differentiating both sides with respect to $x$ using the chain rule gives:

$$\begin{array}{rl}1& =\frac{d}{dx}(x)\\ & =\frac{d}{dx}f(y)\\ & =\frac{df}{dy}\times \frac{dy}{dx}\\ & =\frac{dx}{dy}\times \frac{dy}{dx}\end{array}$$

Hence, if $\frac{dx}{dy}\ne 0:$

$$\frac{dy}{dx}=\frac{1}{\left(\frac{dx}{dy}\right)}$$

6.6 L’Hopital’s Rule

Suppose that $f$ and $g$ are differentiable and $g^{\prime }(x)\ne 0$ near $a$ (except possibly at $a$). Suppose that

$$\underset{x\to a}{\lim }f(x)=0\text{ and }\underset{x\to a}{\lim }g(x)=0$$

or

$$\underset{x\to a}{\lim }f(x)=\pm \infty \text{ and }\underset{x\to a}{\lim }g(x)=\pm \infty$$

then

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

if the limit on the right exists or is $\pm \infty$.

6.6.1 Example

Find ${\lim }_{x\to 1}\frac{\ln x}{x-1}$.

Note that

$$\underset{x\to 1}{\lim }\ln x=\underset{x\to 1}{\lim }(x-1)=0$$

Hence, we cannot use

$$\underset{x\to 0}{\lim }\frac{f(x)}{g(x)}=\frac{{\lim }_{x\to 0}f(x)}{{\lim }_{x\to 0}g(x)}$$

Instead, we use L’Hopital’s rule to get:

$$\underset{x\to 1}{\lim }\frac{\ln x}{x-1}=\underset{x\to 1}{\lim }\frac{\frac{1}{x}}{1}$$

if this limit exits,

$$=1.$$

6.7 Continuous Extension of Sequences

Sometimes L’Hopital’s rule can be used to evaluate limits of sequences. Let $f$ be a function on the real numbers such that ${\lim }_{x\to \infty }f(x)$ exists. Let $f(n)=a_n$ for natural numbers $n$. Then:

$$\underset{n\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }f(x)$$

6.7.1 Example

Evaluate ${\lim }_{n\to \infty }\frac{\ln x}{n}$.

Define $f(x)=\frac{\ln x}{x}$. Hence

$$\begin{array}{rl}\underset{x\to \infty }{\lim }f(x)& =\underset{x\to \infty }{\lim }\frac{\ln x}{x}\\ & =\underset{x\to \infty }{\lim }\frac{1/x}{1}\text{ by L’Hopital’s rule}\\ & ⟹\underset{n\to \infty }{\lim }\frac{\ln n}{n}=0\end{array}$$

6.8 The Mean Value Theorem (MVT)

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then

$$\frac{f(b)-f(a)}{b-a}=f^{\prime }(c)$$

for some $c$, where $a<c<b$.

Note $f^{\prime }(c)$ is the slope of $y=f(x)$ at $x=c$ and $\frac{f(b)-f(a)}{b-a}$ is the slope of the chord joining $A=(a,f(a))$ to $B=(b,f(b))$.

6.9 Increasing/Decreasing Test

Suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$.

8.4 Critical Points and Curve Sketching

A function $f$ has a global maximum at $c$ if $f(c)\ge f(x)$ for all $x$ in the domain of $f$. The number $f(c)$ is called the maximum value of $f$ on its domain. A global maximum is also called an absolute maximum.

A function $f$ has a global minimum at $c$ if if $f(c)\le f(x)$ for all $x$ in the domain of $f$. The number $f(c)$ is called the minimum value of $f$ on its domain. A global minimum is also called an absolute minimum.

A function $f$ has a local maximum at $c$ if $f(c)\ge f(x)$ for all $x$ near $c$.

A function $f$ has a local minimum at $c$ if $f(c)\le f(x)$ for all $x$ near $c$.

If $f$ has a local maximum or minimum at $a$, and if $f^{\prime }(a)$ exists, then $f^{\prime }(a)=0$. The point $(a,f(a))$ is a critical point of the function $f$ if $f^{\prime }(a)=0$ or if $f^{\prime }(a)$ does not exists (but $f(a)$ does).

Thus, all local maxima and minima are critical points. Note however, that not all critical points are local maxima or minima.

To find any local maximum or minimum of a function $f$, we solve the equation $f^{\prime }=0$. We can then classify any critical points we find using the information about the function near the critical point.

A function $f$ is strictly increasing on an interval $[a,b]$ if for all $x_1$ and $x_2$ in $[a,b],f(x_1)<f(x_2)$ whenever $x_1<x_2$.

If $f^{\prime }(x)>0$ on an interval, then $f$ is strictly increasing on that interval. If $f^{\prime }(x)<0$ on an interval, then $f$ is strictly decreasing on that interval. If $f^{\prime }(x)=0$ on an interval, then $f$ is constant on that interval.

First Derivative Test

Suppose that the function $f$ has a critical point at $x=c$. Then

If $f^{\prime }$ changes sign from positive to negative at $c$, then $f$ has a local maximum at $c$. If $f^{\prime }$ changes sign from negative to positive at $c$, then $f$ has a local minimum at $c$. If $f^{\prime }$ does not change sign at $c$, then $f$ has neither a local maximum nor a local minimum at $c$.

The Second Derivative

The second derivative of a function $f$ is the derivative of the derived function $f^{\prime }$. The second derivative of $f$ is denoted $f"$.

The second derivative provides information about the concavity of the graph of a function.

If the graph of $f$ lies above all of its tangent lines on an interval, then it is concave up on that interval. If the graph of $f$ lies below all of its tangent lines on an interval, then it is concave down on that interval.

If $f"(x)>0$ for all $x$ in an interval, then the graph of $f$ is concave up on that interval. If $f"(x)<0$ for all $x$ in an interval, then the graph of $f$ is concave down on that interval.

Second Derivative Test

Suppose $f"$ is a continuous function near a point $c$.

If $f^{\prime }(c)=0$ and $f"(c)>0$, then $f$ has a local minimum (concave up) at $c$. If $f^{\prime }(c)=0$ and $f"(c)<0$, then $f$ has a local maximum (concave down) at $c$. If $f"(c)=0$, then this test is inconclusive.

Curve Sketching

To sketch the curve of a function $y=f(x)$:

• Determine the domain of $f$.
• Determine the $y$-intercept of the graph by evaluating $f(0)$.
• If it is possible to solve the equation $f(x)=0$, find the $x$-intercepts of the graph.
• Determine $f^{\prime }$ and identify the intervals of which $f$ is increasing and the intervals on which $f$ is decreasing.
• Find the critical points of $f$. Determine which critical points are local maxima or local minima (use first or second derivative test).
• Determine $f"$ and identify the intervals on which $f$ is concave up and the intervals on which $f$ is concave down.
• Sketch the graph.

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6.11 Local Maxima and Minima

A function $f$ has a local maximum at $a$ if

$$f(a)\ge f(x)$$

for all $x$ in some open interval containing $a$.

Similarly, $f$ has local minimum at $b$ if

$$f(b)\ge f(x)$$

for all $x$ in some open interval containing $b$.

6.11.1 Critical Points

A function $f$ is said to have a critical point at $x=a$, $a\in \text{dom}(f)$ if $f^{\prime }(a)=0$ or if $f^{\prime }(a)$ does not exist.

6.11.2 Global Maximum and Minimum

A function $f$ has a global maximum at $c$ if $f(c)\ge f(x)$ for all $x$ in the domain of $f$. The number $f(c)$ is called the maximum value of $f$ on its domain. A global maximum is also called an absolute maximum.

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A function $f$ has a global minimum at $c$ if if $f(c)\le f(x)$ for all $x$ in the domain of $f$. The number $f(c)$ is called the minimum value of $f$ on its domain. A global minimum is also called an absolute minimum.

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