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## 17Calculus - Derivatives and Differentiation

### Derivatives

Graphing

Related Rates

Optimization

Other Applications

### Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

Calculus Tools

### Articles

This page consists of some basic topics to get you started with derivatives with the limit definition of the derivative. Before we get started on the details of derivatives and differentiation, let's watch this introduction video. This video goes pretty deeply into derivatives so you don't need to understand everything but it is a good place to start.

### Dr Chris Tisdell - Beginners Guide to Differentiation in Calculus

video by Dr Chris Tisdell

Derivative Definition and Meaning

$$\displaystyle{ f~'(x) = }$$ $$\displaystyle{ \lim_{\Delta x \to 0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}} }$$

derivative = slope

you may also see the derivative referred to as the rate of change or instantaneous rate of change

There are two very important things to remember about the derivative, the definition and what it means.
First, the definition of the derivative is from a limit. When you were studying limits, you may have run across this limit but not known what it meant. There are at least three different forms that you might see. These are sometimes called difference quotients.

Three Ways to Write the Limit Definition of the Derivative

$$\displaystyle{ f~'(x) = \lim_{\Delta x \to 0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}} }$$

original form; used the most since it contains the most information

$$\displaystyle{ f~'(x) = \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} }$$

alternative form of the first equation; some students find this form easier to use

$$\displaystyle{ f~'(a) = \lim_{x \to a}{\frac{f(x) - f(a)}{x-a}} }$$

this form is usually used when finding the derivative at a point $$x=a$$; used the least since it is specific to one point

The second thing you need to really know is that the derivative is the slope of the graph. Every time you hear the word 'derivative' you should think 'slope' and, when a problem asks for the slope (or the equation of the tangent line, since it involves slope) your first thought should be 'derivative'.

Okay, so what do all these equations actually mean and how are you supposed to understand them? Here is a fun video, using the secant line, which gives you an intuitive understanding of what the derivative actually is.

### Krista King Math - Secant line [2min-34secs]

video by Krista King Math

Before we discuss some details about the derivative, let's talk about notation.

Derivative Notation

### Why Notation is Important in Calculus

Using correct notation is extremely important in calculus. If you truly understand calculus, you will use correct notation. Take a few extra minutes to notice and understand notation whenever you run across a new concept. Start using correct notation from the very first.

You may not think this is important. However, if your current (or previous) teacher doesn't require correct notation, learn it on your own. You may (and probably will) get a teacher in the future that WILL require correct notation and this will cause you problems if you don't learn it now. It is much easier to learn it correctly the first time than to have to correct your notation later, after you have been doing it incorrectly for a while.

Not only is it important in class to use correct notation, when you use math ( or any other subject that uses special symbols ) in your career, you will need to be able to communicate what you mean. Without correct notation, your ideas could be misunderstood. It's a lot like speaking English ( or whatever language you use regularly ) or speaking a variant that has meaning only to you. You will be misunderstood and it may even affect your ability to keep a job.

So, just decide to start using correct notation now. It's not that hard and it will pay off in the long run.

You will see and use various types of notation to write this derivative. Most books just tell what to write without explaining why. Let me try to explain. In the following examples, I will assume $$y = f(x)$$.

Various Notation Methods to Write the Derivative

$$f~'(x)$$

Used quite often and the small quote symbol indicates the derivative. Make sure that this symbol appears clearly in your work.

$$y'$$

Also used a lot. The disadvantage of this notation is that you need to pick up from the context what the variable of differentiation is, since it is not explictly stated in the expression, like it is in the previous example.

$$\displaystyle{\frac{d}{dx}[f(x)]}$$

Probably the clearest way to write the derivative but is also longer than the previous two ways. Use of this notation is usually limited to situations when it is important to clearly show the function itself and the variable of differentiation.

$$\displaystyle{\frac{dy}{dx}}$$

Also quite common. This is shorthand for $$\displaystyle{\frac{d[y]}{dx} }$$ and $$\displaystyle{ \frac{d}{dx}[y]}$$. [ see note below ]

$$D_x[y]$$ and $$D_x[f(x)]$$

Not used much in beginning calculus. However, they are important when you study partial derivatives.

Note - - The notation $$dy/dx$$ does NOT mean dy divided by dx. It means the derivative of y with respect to x. Right before you begin studying integration, you will probably cover a section about differentials. This section will prove that you can write $$\displaystyle{\frac{dy}{dx} = f~'(x)}$$ in differential form as $$dy = [ f~'(x) ] dx$$. This can propagate the illusion of division but don't let yourself be fooled. It takes quite a bit to prove the differential form.

Using The Limit Definition

Okay, let's watch some videos related to the derivative and how to use the limit definition to get our heads around this concept.

This first video contains a great explanation of how to think about this limit definition and what it means. It also contains a good example of how to calculate and interpret the limit.

### Krista King Math - Derivatives [7min-48secs]

video by Krista King Math

After watching that first video, if the ideas are still not clear, here is another good video for you to watch explaining the same concepts.

### PatrickJMT - Understanding the Definition of the Derivative [12min-8secs

video by PatrickJMT

Here is video that compares the graph of a function to the graph it's derivative. It is a great follow-up to the previous video.

### PatrickJMT - Sketching the Derivative of a Function [14min-49secs]

video by PatrickJMT

Fortunately, we are not going to have to use this limit definition to calculate derivatives all the time. There are some basic rules, based on the limit definition, that will very quickly replace the use the limit definition to calculate derivatives.

Coming Up

Here is a great video with quick explanations of the next 4 or 5 rules that you will learn to calculate derivatives. This video is especially helpful if you are refreshing your calculus skills or just need a review before an exam. If you are first learning derivatives, it is better to skip this video for now and come back to it right before your first exam on derivatives.

### Krista King Math - Solving Derivatives [9min-0secs]

video by Krista King Math

After working some practice problems, your next logical step is to learn a few basic rules on the constant and addition rules page

Practice

Unless otherwise instructed, use the limit definition to calculate the derivative in these problems. Give your answers in exact terms, completely factored.

Basic

$$f(x) = 12 + 7x$$

Problem Statement

Use the limit definition to calculate the derivative of $$f(x) = 12 + 7x$$

$$f'(x) = 7$$

Problem Statement

Use the limit definition to calculate the derivative of $$f(x) = 12 + 7x$$

Solution

From the problem statement, we are given $$f(x)=12+7x$$
We will use the equation $$\displaystyle{ f'(x) = \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} }$$ but the equation $$\displaystyle{ f'(x) = \lim_{\Delta x \to 0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}} }$$ will also work.
In our equation of choice, we have two main factors, $$f(x)$$ and $$f(x+h)$$. We are given $$f(x)=12+7x$$. To get $$f(x+h)$$ we replace x with x+h everywhere in $$f(x)$$. This gives us $$f(x+h) = 12+7(x+h)$$. Notice that there is only one x here, so we just need to do the substitution once. If there was more than one place that an x appears, we would need to replace EVERY x with x+h.
Okay, so now we are ready to set up the limit.

 $$\displaystyle{ f'(x) = \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} = \lim_{h\to0}\frac{[12+7(x+h)]-(12+7x)}{h}}$$ Now we just use algebra to simplify and solve. $$\displaystyle{ f'(x) = \lim_{h\to0}\frac{12+7x +7h-12-7x}{h} }$$ $$\displaystyle{ f'(x) = \lim_{h\to0}\frac{7h}{h} }$$ $$\displaystyle{ f'(x) = \lim_{h\to0}7 = 7 }$$

This is also solved in this video.

### 899 video

video by PatrickJMT

$$f'(x) = 7$$

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$$f(x) = x^2 + x - 3$$

Problem Statement

Use the limit definition to calculate the derivative of $$f(x) = x^2 + x - 3$$

$$f'(x) = 2x + 1$$

Problem Statement

Use the limit definition to calculate the derivative of $$f(x) = x^2 + x - 3$$

Solution

### 901 video

video by PatrickJMT

$$f'(x) = 2x + 1$$

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Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x = 3$$ for $$\displaystyle{ f(x) = x^2 + 1 }$$

Problem Statement

Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x = 3$$ for $$\displaystyle{ f(x) = x^2 + 1 }$$

Solution

### 3688 video

video by rootmath

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$$f(x) = x^2 + 5$$

Problem Statement

Use the limit definition to calculate the derivative of $$f(x) = x^2 + 5$$

$$f'(x) = 2x$$

Problem Statement

Use the limit definition to calculate the derivative of $$f(x) = x^2 + 5$$

Solution

### 905 video

video by Krista King Math

$$f'(x) = 2x$$

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Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x = 2$$ for $$\displaystyle{ f(x) = 3x^2+5 }$$

Problem Statement

Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x = 2$$ for $$\displaystyle{ f(x) = 3x^2+5 }$$

Solution

### 3696 video

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Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x = 1$$ for $$\displaystyle{ f(x) = \sqrt{x} }$$

Problem Statement

Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x = 1$$ for $$\displaystyle{ f(x) = \sqrt{x} }$$

Solution

### 3689 video

video by rootmath

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$$\displaystyle{f(x)=\frac{\sqrt{x}}{8}}$$

Problem Statement

Use the limit definition to calculate the derivative of $$\displaystyle{f(x)=\frac{\sqrt{x}}{8}}$$

$$\displaystyle{ f'(x)=\frac{1}{16\sqrt{x}}}$$

Problem Statement

Use the limit definition to calculate the derivative of $$\displaystyle{f(x)=\frac{\sqrt{x}}{8}}$$

Solution

In the video solution below, he uses a shortcut that you are not usually given to solve this problem. To solve this problem without the shortcut, you need to use the concept of rationalizing that we discuss on the finite limits page.

 $$\displaystyle{\frac{1}{8}\lim_{h \to 0}{\left[\frac{\sqrt{x+h}-\sqrt{h}}{h}\right] }}$$ $$\displaystyle{ \frac{\sqrt{x+h}+\sqrt{h}}{\sqrt{x+h}+\sqrt{h}}}$$ $$\displaystyle{ \frac{1}{8}\lim_{h \to 0}{ \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} } = \frac{1}{8} \left[ \frac{1}{2\sqrt{x}} \right] = \frac{1}{16\sqrt{x}}}$$

### 906 video

video by PatrickJMT

$$\displaystyle{ f'(x)=\frac{1}{16\sqrt{x}}}$$

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Intermediate

$$\displaystyle{ f(x) = \frac{1}{1+x^2} }$$

Problem Statement

Use the limit definition to calculate the derivative of $$\displaystyle{ f(x) = \frac{1}{1+x^2} }$$

Solution

### 898 video

video by MIT OCW

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$$\displaystyle{ G(t) = \frac{4t}{t+1} }$$

Problem Statement

Use the limit definition to calculate the derivative of $$\displaystyle{ G(t) = \frac{4t}{t+1} }$$

Solution

### 900 video

video by PatrickJMT

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$$\displaystyle{ f(x) = \frac{x}{1-2x} }$$

Problem Statement

Use the limit definition to calculate the derivative of $$\displaystyle{ f(x) = \frac{x}{1-2x} }$$

Solution

### 907 video

video by Krista King Math

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Use the limit definition to calculate $$f'(x)$$ for $$f(x)= 3/(4x)$$ and find the slope of the tangent line at $$x=2$$.

Problem Statement

Use the limit definition to calculate $$f'(x)$$ for $$f(x)= 3/(4x)$$ and find the slope of the tangent line at $$x=2$$.

Solution

### 902 video

video by PatrickJMT

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Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x=4$$ for $$\displaystyle{f(x)=\sqrt{2x+1}}$$

Problem Statement

Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x=4$$ for $$\displaystyle{f(x)=\sqrt{2x+1}}$$

Solution

### 903 video

video by PatrickJMT

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Use the limit definition to calculate the slope and equation of the tangent line to the function $$f(x)=\sqrt{x+2}$$ at the point $$(7,3)$$

Problem Statement

Use the limit definition to calculate the slope and equation of the tangent line to the function $$f(x)=\sqrt{x+2}$$ at the point $$(7,3)$$

Solution

### 1312 video

video by PatrickJMT

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$$\displaystyle{ f(x) = \frac{x}{x+3} }$$

Problem Statement

Use the limit definition to calculate the derivative of $$\displaystyle{ f(x) = \frac{x}{x+3} }$$

Solution

### 908 video

video by Krista King Math

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Find the equation of the tangent line to the graph of $$f(x)=x^2-3$$ at $$(3,6)$$ using the limit definition of the derivative.

Problem Statement

Find the equation of the tangent line to the graph of $$f(x)=x^2-3$$ at $$(3,6)$$ using the limit definition of the derivative.

$$f(x)=x^2-3$$

Problem Statement

Find the equation of the tangent line to the graph of $$f(x)=x^2-3$$ at $$(3,6)$$ using the limit definition of the derivative.

Solution

### 2202 video

video by PatrickJMT

$$f(x)=x^2-3$$

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Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x = 4$$ for $$\displaystyle{ f(x) = 1/x }$$

Problem Statement

Use the limit definition to calculate $$f'(x)$$ and the slope of the tangent line at $$x = 4$$ for $$\displaystyle{ f(x) = 1/x }$$

Solution

### 3690 video

video by rootmath

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Use the limit definition to determine the point on the graph of $$\displaystyle{y=1/x+2}$$ where the slope is equal to $$-2$$.

Problem Statement

Use the limit definition to determine the point on the graph of $$\displaystyle{y=1/x+2}$$ where the slope is equal to $$-2$$.

Solution

If you were confused by the equation in the question whether it meant $$(1/x)+2$$ or $$1/(x+2)$$, you need to review order of operations in the precalculus section.

### 904 video

video by PatrickJMT

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You CAN Ace Calculus

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

### Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

Engineering

Circuits

Semiconductors

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 Derivative Definition and Meaning Derivative Notation Using The Limit Definition Coming Up Practice

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# Maharashtra Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.1

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 6 Definite Integration Ex 6.1 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.1

Evaluate the following definite integrals:

Question 1.
$$\int_{4}^{9} \frac{1}{\sqrt{x}} d x$$
Solution:

Question 2.
$$\int_{-2}^{3} \frac{1}{x+5} d x$$
Solution:

Question 3.
$$\int_{2}^{3} \frac{x}{x^{2}-1} d x$$
Solution:

Question 4.
$$\int_{0}^{1} \frac{x^{2}+3 x+2}{\sqrt{x}} d x$$
Solution:

Question 5.
$$\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x$$
Solution:
Let I = $$\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x$$
Let $$\frac{x}{(x+2)(x+3)}=\frac{A}{x+3}+\frac{B}{x+2}$$
∴ x = A(x + 2) + B(x + 3)
Put x + 3 = 0, i.e. x = -3, we get
-3 = A(-1) + B(0)
∴ A = 3
Put x + 2 = 0, i.e. x = -2, we get
-2 = A(0) + B(1)
∴ B = -2

Question 6.
$$\int_{1}^{2} \frac{d x}{x^{2}+6 x+5}$$
Solution:

Question 7.
If $$\int_{0}^{a}(2 x+1) d x$$ = 2, find the real values of ‘a’.
Solution:
Let I = $$\int_{0}^{a}(2 x+1) d x$$
= $$\left[2 \cdot \frac{x^{2}}{2}+x\right]_{0}^{a}$$
= a2 + a – 0
= a2 + a
∴ I = 2 gives a2 + a = 2
∴ a2 + a – 2 = 0
∴ (a + 2)(a – 1) = 0 1
∴ a + 2 = 0 or a – 1 = 0
∴ a = -2 or a = 1.

Question 8.
If $$\int_{1}^{a}\left(3 x^{2}+2 x+1\right) d x$$ = 11, find ‘a’.
Solution:
Let I = $$\int_{1}^{a}\left(3 x^{2}+2 x+1\right) d x$$
= $$\left[3\left(\frac{x^{3}}{3}\right)+2\left(\frac{x^{2}}{2}\right)+x\right]_{1}^{a}$$
= $$\left[x^{3}+x^{2}+x\right]_{1}^{a}$$
= (a3 + a2 + a) – (1 + 1 + 1)
= a3 + a2 + a – 3
∴ I = 11 gives a3 + a2 + a – 3 = 11
∴ a3 + a2 + a – 14 = 0
∴ (a3 – 8) + (a2 + a – 6) = 0
∴ (a – 2)(a2 + 2a + 4) + (a + 3)(a – 2) = 0
∴ (a – 2)(a2 + 2a + 4 + a + 3) = 0
∴ (a – 2)(a2 + 3a + 7) = 0
∴ a – 2 = 0 or a2 + 3a + 7 = 0
∴ a = 2 or a = $$\frac{-3 \pm \sqrt{9-28}}{2}$$
The latter two roots are not real.
∴ they are rejected.
∴ a = 2.

Question 9.
$$\int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} d x$$
Solution:

Question 10.
$$\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x$$
Solution:
Let I = $$\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x$$ = $$\int_{1}^{2} \frac{3 x}{(3 x)^{2}-1} d x$$
Put 3x = t
∴ 3 dx = dt
∴ dx = $$\frac{d t}{3}$$
When x = 1, t = 3 × 1 = 3
When x = 2, t = 3 × 2 = 6

Question 11.
$$\int_{1}^{3} \log x d x$$
Solution:

format: plain


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# Critical nonlinear Schrödinger equations in higher space dimensions

## Abstract

We study the critical nonlinear Schrödinger equations $i\partial _{t}u+\frac{1}{2}\Delta u = \lambda \vert u\vert^{2/n}u, \quad (t,x) \in \mathbb{R}^{+}\times \mathbb{R}^{n},$ in space dimensions 𝑛 ≥ 4, where 𝜆 ∈ ℝ. We prove the global in time existence of solutions to the Cauchy problem under the assumption that the absolute value of Fourier transform of the initial data is bounded below by a positive constant. Also we prove the two side sharp time decay estimates of solutions in the uniform norm.

## Journal

• Journal of the Mathematical Society of Japan

Journal of the Mathematical Society of Japan 70(4), 1475-1492, 2018

The Mathematical Society of Japan

## Codes

• NII Article ID (NAID)
130007500591
• NII NACSIS-CAT ID (NCID)
AA0070177X
• Text Lang
ENG
• ISSN
0025-5645
• NDL Article ID
029291992
• NDL Call No.
Z53-A209
• Data Source
NDL  J-STAGE

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