Deriving a Function Without Quotient Rule

  • Thread starter Arnoldjavs3
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    Derivative
In summary, the student is trying to derive the function f(x) without using the quotient or product rule. After reviewing basic algebra, they are able to solve for f(x) using the distributive property.
  • #1
Arnoldjavs3
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Poster has been reminded to show more work when posting schoolwork questions

Homework Statement


derive this function

f(x) = (2x + x^3) / sqrt(x)

Homework Equations

The Attempt at a Solution


how can i derive this without using quotient rule? my prof is asking to do so without using it.

I
 
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  • #2
Can you rewrite it in a way that would allow you to use the product rule?
 
  • #3
Arnoldjavs3 said:

Homework Statement


derive this function

f(x) = (2x + x^3) / sqrt(x)

Homework Equations

The Attempt at a Solution


how can i derive this without using quotient rule? my prof is asking to do so without using it.

I
Write ##\ \sqrt{x} \ ## as ##\displaystyle \ x^{1/2} \ ## .By the way, the word in English is differentiate, not derive.
 
  • #4
Can this be done without product / quotient rule?

Just to remind myself of how quotient rule works...

I got to this point after using quotient rule:
((2x + 7x^3) / (2(x^1/2))) / x

How do i simplify this? Sorry I don't know how to use latex code.
 
  • #5
OK, here's a hint for how to do it without the product rule either: ##x^m / x^n = x^{(m-n)} ##.
 
  • #6
Okay using that I got something like x^5/2 + x^1/2. It doesn't seem correct, how would i account for the binomial on the numerator? would I write another expression for x such as x^h?
\
edit: I just misunderstood the question... I am allowed to use both the quotient and product rule.
 
  • #7
Arnoldjavs3 said:
Okay using that I got something like x^5/2 + x^1/2. It doesn't seem correct, how would i account for the binomial on the numerator?
It's not correct, but you're not too far off. And you should end up with two terms, so I don't understand what you're asking about accounting for the binomial.
Arnoldjavs3 said:
would I write another expression for x such as x^h?
What do you get if you carry out the division below?
(2x + x^3) / sqrt(x)
Arnoldjavs3 said:
edit: I just misunderstood the question... I am allowed to use both the quotient and product rule.
 
  • #8
alright... I got it.

I simply just subtract the exponents from the numerator and denominator so it became (2 + 5x^2) / 2x^1/2
 
  • #9
Arnoldjavs3 said:
alright... I got it.

I simply just subtract the exponents from the numerator and denominator so it became (2 + 5x^2) / 2x^1/2
No, that's not even close. There's some very basic algebra that you need to review. If you don't, you absolutely won't be able to do calculus.

##\frac{(a + b)} c = (a + b) \cdot \frac 1 c = a \cdot \frac 1 c + b \cdot \frac 1 c## . Use the distributive property to multiply each term of a + b by 1/c. Can you apply this idea to your problem, ##\frac{2x + x^3}{\sqrt{x}}##?
 

1. What is a derivative?

A derivative is a mathematical concept that measures the rate of change of one quantity with respect to another. It is often referred to as the slope of a curve at a given point.

2. How do you find the derivative of a function?

The derivative of a function is found by taking the limit of the average rate of change as the interval between two points approaches zero. This can be done using the power rule, product rule, quotient rule, or chain rule, depending on the complexity of the function.

3. What is the purpose of finding a derivative?

Finding a derivative is useful in many areas of science, such as physics, engineering, and economics. It allows us to understand how a system changes over time and to make predictions about its future behavior.

4. Can you give an example of a real-world application of derivatives?

One example of a real-world application of derivatives is in calculating the velocity of an object at a particular point in time. By finding the derivative of the object's position function, we can determine its instantaneous velocity at that moment.

5. Are there any limitations to using derivatives?

While derivatives are a powerful tool in mathematics and science, there are some limitations to their use. For example, they may not be able to accurately describe systems with abrupt changes or discontinuities. Additionally, they may not be appropriate for nonlinear or chaotic systems.

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