Calculus Homework: Derivative of g(x) = x * sqrt(4-x) using Product Rule

In summary, the conversation discusses finding the derivative of g(x)=x * sqrt(4-x) using the product rule and rewriting the function as (4-x)^1/2. The final solution is (x/2 * sqrt(4-x)) + sqrt(4-x).
  • #1
Chaubin
4
0

Homework Statement


Find the derivative of g(x)=x * sqrt(4-x)

Homework Equations


The Attempt at a Solution


I know I use the product rule, but I am not sure how to derive the sqrt(4-x) portion.
 
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  • #2
Welcome to PF!

Hi Chaubin! Welcome to PF! :smile:

(have a square-root: √ :wink:)
Chaubin said:
… I am not sure how to derive the sqrt(4-x) portion.

Hint: rewrite it as (4 - x)1/2 :smile:
 
  • #3
What's another way to write sqrt(4-x), say... with exponents?
 
  • #4
So, I take the quantity (4-x)^1/2 and turn it into (1/2)*(4-x)^-1/2? Is that it?
 
  • #5
Chaubin said:
So, I take the quantity (4-x)^1/2 and turn it into (1/2)*(4-x)^-1/2? Is that it?

Remember the chain rule.
 
  • #6
(try using the X2 icon just above the Reply box :wink:)

you've left out the coefficient of x :redface:
 
  • #7
Ok, i think I finished it.
I ended up with (x/((2 * sqrt(4-x))) + sqrt(4-x)
 
  • #8
Chaubin said:
Ok, i think I finished it.
I ended up with (x/((2 * sqrt(4-x))) + sqrt(4-x)

no, that would work for √(x-4), but not for √(4-x)

(and your brackets are in the wrong place)
 
  • #9
I know the derivative of sqrt(4-x) is -1/(2*sqrt(4-x)).
After this I get a bit foggy on the procedure.
I think I have to use the product rule f(x)*g(x)= f(x)*g'(x) + f'(x)*g(x)
In this case f(x) = x and g(x)= sqrt(4-x).
So it should be:
x*(-1/(2*sqrt(4-x)) + 1 * sqrt(4-x)

correct?
 
  • #10
Hi Chaubin! :smile:

(just got up :zzz: …)
Chaubin said:
x*(-1/(2*sqrt(4-x)) + 1 * sqrt(4-x)

correct?

Yup! :biggrin:
 

1. What is a derivative in calculus?

A derivative in calculus is a way to measure the rate of change of a function. It is the slope of the tangent line at a specific point on a curve. In simpler terms, it tells us how much a function is changing at a certain point.

2. Why is it important to know how to find derivatives?

Derivatives are important in many areas of science and engineering, as well as in everyday life. They allow us to analyze and model the behavior of complex systems, such as the movement of objects, the growth of populations, and the change in temperature over time. They also have many practical applications, such as optimizing functions and solving optimization problems.

3. Can you explain the process of finding a derivative?

The process of finding a derivative involves using specific formulas and rules to determine the slope of the tangent line at a given point on a curve. This can be done using the limit definition of a derivative or by using rules such as the power rule, product rule, and chain rule. It is important to have a good understanding of algebra and trigonometry when finding derivatives.

4. What are some real-life examples of how derivatives are used?

Derivatives are used in many fields, such as physics, engineering, economics, and finance. In physics, derivatives are used to calculate the velocity and acceleration of objects. In economics, derivatives are used to analyze the rate of change of supply and demand curves. In finance, derivatives are used to evaluate the risk and return of investments.

5. Are there any common mistakes to avoid when finding derivatives?

One common mistake when finding derivatives is forgetting to apply the chain rule or product rule correctly. It is important to carefully follow the rules and formulas and to pay attention to the order of operations. Another mistake is not simplifying the final answer, which can lead to incorrect solutions. It is also important to practice and check your work to avoid any calculation errors.

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