I have problems with differentiating

In summary, Homework Equations are H(x)=(x+x^-1)^3, but for the first one I got x^3/2 + 4x^1/2 + 3x^-1/2 but I don't know if that is correct or where to go from here? and the second I really don't know what to do, I am not sure if I have to distribute the exponent 3 throughout the parenthesis? I really don't know what to do, if someone could give me a hint? then I could work it out. A friend of mine got x^3 + 3x^2(x^-1) + 3x(x^-2) + x^-
  • #1
Altami
17
0

Homework Statement


Differentiate.

y=x^2 + 4x +3/(square root x)

and

H(x)=(x+x^-1)^3


Homework Equations





The Attempt at a Solution



For the first one I got x^3/2 + 4x^1/2 + 3x^-1/2
But I don't know if that is correct or where to go from here?

The second I really don't know what to do, I am not sure if I have to distribute the exponent 3 throughout the parenthesis? I really don't know what to do, if someone could give me a hint? Then I could work it out. A friend of mine got x^3 + 3x^2(x^-1) + 3x(x^-2) + x^-3. But I don't know how he got that? Could someone explain if he is right or how he got that?

Thanks.
 
Physics news on Phys.org
  • #2


Altami said:

Homework Statement


Differentiate.

y=x^2 + 4x +3/(square root x)

and

H(x)=(x+x^-1)^3


Homework Equations





The Attempt at a Solution



For the first one I got x^3/2 + 4x^1/2 + 3x^-1/2
But I don't know if that is correct or where to go from here?

The second I really don't know what to do, I am not sure if I have to distribute the exponent 3 throughout the parenthesis? I really don't know what to do, if someone could give me a hint? Then I could work it out. A friend of mine got x^3 + 3x^2(x^-1) + 3x(x^-2) + x^-3. But I don't know how he got that? Could someone explain if he is right or how he got that?

Thanks.
Your answer to the first one is not correct. You need to differentiate x^2, 4x and 3/x^(1/2) independently and then add the stuff together. To find the derivative of 3/(x^(1/2)), it is easier to write it this way: 3x^(-1/2).

As for the second one, I haven't worked it out, but you have two options for how to do this 1)write it as: (x+x^-1)(x+x^-1)(x+x^-1) and find the derivative (this is going to get ugly)

2)Use the chain rule
 
  • #3


There's another option on the third one: expand it and then take the derivative. The cube of a binomial isn't too hard.
(a + b)3 = a3 + 3a2b + 3ab2 + b3.

You can apply this formula to (x + 1/x)3. After expansion, just take the derivative of the four terms.
 
  • #4


Mark44 said:
There's another option on the third one: expand it and then take the derivative. The cube of a binomial isn't too hard.
(a + b)3 = a3 + 3a2b + 3ab2 + b3.

You can apply this formula to (x + 1/x)3. After expansion, just take the derivative of the four terms.

Ahh yes, Mark's way is better.
 
  • #5


Okay let me try it and I'll show my answer hopefully I'll get the right one, thanks guys.

Okay for the second one I got

H'(x)=3x^2 + 3 - 3x^-2 - 3x^-4...right?

And for the first one I got

3/2x^1/2 + 2x^-1/2 - -3/2x^-3/2
 
Last edited:
  • #6


Altami said:
Okay let me try it and I'll show my answer hopefully I'll get the right one, thanks guys.

Okay for the second one I got

H'(x)=3x^2 + 3 - 3x^-2 - 3x^-4...right?
Yes, right.
Altami said:
And for the first one I got

3/2x^1/2 + 2x^-1/2 - -3/2x^-3/2
I don't know. This is what you wrote:
Altami said:
y=x^2 + 4x +3/(square root x)

From your work, it looks like you meant this.
[tex]y = \frac{x^2 + 4x + 3}{\sqrt{x}}[/tex]

If you intended that all three terms should be in the numerator, put parentheses around the entire numerator, like this: y = (x2 + 4x + 3)/x1/2

So which was your problem?

Also, exponents that consist of more than a single digit, use parentheses, as in x^(1/2), x^(-1/2), and so on.
 
  • #7


Mark44 said:
Yes, right.

I don't know. This is what you wrote:


From your work, it looks like you meant this.
[tex]y = \frac{x^2 + 4x + 3}{\sqrt{x}}[/tex]

If you intended that all three terms should be in the numerator, put parentheses around the entire numerator, like this: y = (x2 + 4x + 3)/x1/2

So which was your problem?

Also, exponents that consist of more than a single digit, use parentheses, as in x^(1/2), x^(-1/2), and so on.

Oh sorry, yea it is...[tex]y = \frac{x^2 + 4x + 3}{\sqrt{x}}[/tex]
 
  • #8


The first two terms in your answer are correct, but the third one isn't.

That part of your answer is also ambiguous.
3/2x^-3/2

Does this mean
[tex]\frac{3}{2x^{-3/2}}[/tex]

or
[tex]\frac{3}{2}x^{-3/2}[/tex]
?
 
  • #9


Mark44 said:
The first two terms in your answer are correct, but the third one isn't.

That part of your answer is also ambiguous.
3/2x^-3/2

Does this mean
[tex]\frac{3}{2x^{-3/2}}[/tex]

or
[tex]\frac{3}{2}x^{-3/2}[/tex]
?

[tex]\frac{3}{2}x^{-3/2}[/tex]

Sorry I need to start using the Latex I know...
 
  • #10


It's not really that hard (LaTeX). You can click an expression built from LaTeX to see how it was coded.
 
  • #11


Mark44 said:
It's not really that hard (LaTeX). You can click an expression built from LaTeX to see how it was coded.

Yea I actually downloaded the PDF, just that I haven't come on these boards in a while and I kind of forget the coding...:redface:, but thanks for your help.
 
  • #12


Unless you'll be asking a few questions, for the one time off you don't need to learn LaTeX. Just be sure to use brackets as you would have done in the older calculators.

For example, rather than 3/2x^-3/2 you should be writing (3/2)x^(-3/2).

Back to your questions, for y=(x^2 + 4x +3)/sqrt(x) you can split this up into y=x^2/sqrt(x)+4x/sqrt(x)+3/sqrt(x) and change the powers of x accordingly, then differentiate.

For the second question, the chain rule would be easiest imo. Expanding can be done too, but it leaves more room for errors to be made and you'll probably be given higher powers in your exams and expanding will just be out of the question - so it's best if you become sufficient at using the chain rule now.
 

What is differentiation?

Differentiation is a mathematical process of finding the rate of change of a function with respect to its independent variable. It is used to determine the slope of a line or the instantaneous rate of change of a curve at a specific point.

Why do I need to differentiate?

Differentiation is a fundamental concept in calculus and is essential for solving problems related to rates of change, optimization, and curve analysis. It is also used in physics, economics, and other fields to model real-world situations.

What are some common techniques for differentiating?

The most common techniques for differentiation include the power rule, product rule, quotient rule, and chain rule. These rules help to simplify the process of finding the derivative of more complex functions.

What are the applications of differentiation?

Differentiation has a wide range of applications in various fields, including physics, engineering, economics, and biology. It is used to solve problems related to optimization, motion, rates of change, and growth/decay.

How can I improve my differentiation skills?

To improve your differentiation skills, it is important to practice solving various types of differentiation problems and understand the underlying concepts. You can also seek help from a tutor or use online resources to learn and practice different techniques for differentiation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
25
Views
343
  • Calculus and Beyond Homework Help
Replies
10
Views
439
  • Calculus and Beyond Homework Help
Replies
5
Views
783
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
874
  • Calculus and Beyond Homework Help
Replies
3
Views
952
  • Calculus and Beyond Homework Help
Replies
7
Views
553
  • Calculus and Beyond Homework Help
Replies
3
Views
818
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Replies
7
Views
519
Back
Top