Finding the derivative of a function

In summary, we see a student struggling with the chain rule and asking for help in finding the derivative of a given equation. The conversation involves a discussion of using the chain rule and applying the distributive property to simplify the expression. The student is unsure of the correct method and seeks clarification from their professor.
  • #1
meeklobraca
189
0

Homework Statement



y=(5-x2)sqrt(x)



Homework Equations





The Attempt at a Solution



I got as far as

(5-x2)(1/2)(x)^-1/2 + (x)^1/2 (-x)(-2x)

Im using the chain rule here I believe but I am a tad lost as where to go from here, if I've even done it correctly up to this point. Any help would be greatly appreciated.

Thanks!
 
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  • #2
meeklobraca said:
(x)^1/2 (-x)(-2x)

How do you get this? The derivative of x2 is 2x. So it should be [tex]\sqrt{x}[/tex](-2x).
 
  • #3
Im not sure, this chain rule is explained extremely poorly in my manual. Isnt the equatrion I should be using

u * dv/dx + v * du/dx ?
 
  • #4
Yes.
u=(5-x2)
v=[tex]\sqrt{x}[/tex]
Now u dv/dx = (5-x2) ( 1/2 x^-1/2), which you've got,
and v du/dx = [tex]\sqrt{x}[/tex] (-2x).

Add the two parts to get dy/dx.
 
  • #5
I don't know where to go from there.

I got (1/2x^-1/2)(5-x2)

But that seems wrong.
 
  • #6
meeklobraca said:
I got (1/2x^-1/2)(5-x2)

That's just one part. Add u dv/dx and v du/dx to get dy/dx. Then simplify the expression.
 
  • #7
Thats what I thought I did. I had

1/2x^-1/2 (5 - x2 + 2x - 2x) which led me to my answer.
 
  • #8
meeklobraca said:
2x - 2x

How did you get this?
I get,
dy/dx = (5-x2)(1/(2[tex]\sqrt{x}[/tex])) - 2x3/2
dy/dx = 5/(2[tex]\sqrt{x}[/tex]) - 1/2x3/2 -2x3/2

Simplify.
 
  • #9
I don't get how you get to that second line of dy/dx. how did 2 factors minus 1 factor = 3 factors all subtracted from each other. In fact, how you got -2x^3/2 isn't clear either.
 
Last edited:
  • #10
shramana said:
How did you get this?
I get,
dy/dx = (5-x2)(1/(2[tex]\sqrt{x}[/tex])) - 2x3/2
dy/dx = 5/(2[tex]\sqrt{x}[/tex]) - 1/2x3/2 -2x3/2

Simplify.

meeklobraca said:
I don't get how you get to that second line of dy/dx. how did 2 factors minus 1 factor = 3 factors all subtracted from each other. In fact, how you got -2x^3/2 isn't clear either.
The "distributive law": a(b+ c)= ab+ bc.

[tex](5- x^2)(1/(2\sqrt{x})= 5(1/(2\sqrt{x})- x^2(1/2\sqrt{x})[/tex]

Further, [itex]\sqrt{x}= x^{1/2}[/itex] so [itex]1/(2/sqrt{x})= (1/2)x^{-1/2}[/itex] and [itex]x^2(1/2\sqrt{x})= (1/2)x^{2- 1/2}= (1/2)x^{3/2}[/itex].
 
  • #11
We haven't been taught that method, I imagine I won't get marks for it if I do it that way. I have to use one of the derivative rules to get the final answer.
 
  • #12
Which method are you reffering to? Halls' use of the distributive property or the use of the product rule (d(uv)/dx = u'v + v'u) and power rule (d(x^n) = nx^(n-1))?
 
  • #13
The distributive property
 
  • #14
If you've had an elementary algebra course you should be familiar with the distributive property. I don't think you would lose marks for applying an algebraic property to simplify your answers.
 
  • #15
From what I've gathered from the course I think I would. The professor wants an answer to the question using the rules in the unit. There must be an answer to the question using the deriviative rules isn't there?
 
  • #16
Yes, but it's not in it's simplest form. I would clarify with your professor that he doesn't want you to apply algebraic rules to simplify expressions.
 
  • #17
meeklobraca said:
From what I've gathered from the course I think I would. The professor wants an answer to the question using the rules in the unit. There must be an answer to the question using the deriviative rules isn't there?
It was not a question of "derivative rules", it is a question of using algebra. I can't believe that you are not "allowed" to use what you have learned before this course. You are allowed to use arithmetic aren't you?
 

What is the definition of a derivative?

A derivative is a mathematical concept that measures the rate of change of a function at a specific point. It represents the slope of the tangent line to the function at that point.

How do you find the derivative of a function?

To find the derivative of a function, you can use the process of differentiation. This involves using rules and formulas to calculate the derivative of each term in the function, and then simplifying the result to get the final derivative.

What is the purpose of finding the derivative of a function?

The derivative of a function is useful in many areas of mathematics and science. It can be used to find the maximum and minimum values of a function, determine the rate of change of a quantity, and solve optimization problems.

What are the different methods for finding the derivative of a function?

There are several methods for finding the derivative of a function, including the power rule, product rule, quotient rule, and chain rule. Each method is used to find the derivative of a specific type of function.

Can the derivative of a function be negative?

Yes, the derivative of a function can be negative. This indicates that the function is decreasing at that point. A positive derivative indicates that the function is increasing at that point, and a derivative of zero means the function is neither increasing nor decreasing.

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