Derivative Help: t^2[2\sqrt{t} - 1/\sqrt{t}]

  • Thread starter Draggu
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In summary, the conversation is about a student struggling with a calculus problem involving the expression (1/2)t^2[(2\sqrt{t})-(1/\sqrt{t})], which they have tried to solve by expanding and using the product rule for derivatives. However, their answer does not match the correct answer provided on the sheet, which is [(\sqrt{t}(10t-3)/(4)]. Another person points out that the student is missing a t in their second term and suggests differentiating term by term instead of using the product rule. The student asks for further guidance and clarification.
  • #1
Draggu
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Homework Statement


(1/2)t^2[(2[tex]\sqrt{t}[/tex])-(1/[tex]\sqrt{t}[/tex])]

Homework Equations


The Attempt at a Solution



No matter what I do I get the wrong answer
Things I have tried:
-Expanding the problem, then with the derivatives use the product rule (f)(g') + (g)(f')
 
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  • #2
Let's see what you did. Then we can point out where you're going wrong.
 
  • #3
=(1/2)t^2[(2t^(1/2)) - (t^(-1/2))]
=t^(5/2) - (1/2)^(3/2)
Use product rule

= ((5/2)t^(3/2))*((-1/2t)^(3/2)) + ((-3/4t)^(1/2))*(t^(5/2))
=(-5/4)t^3 + (-3/4)t^3
= -7/4t^3

Evidently wrong.

The correct answer on the sheet is [([tex]\sqrt{t}[/tex](10t-3)/(4)]
 
  • #4
Draggu said:
=(1/2)t^2[(2t^(1/2)) - (t^(-1/2))]
=t^(5/2) - (1/2)^(3/2)

That should be [itex]t^{5/2}-\frac{1}{2}t^{3/2}[/itex] (you're missing the [itex]t[/itex] in the second term).

Use product rule

Why? You don't have a product anymore! Just differentiate term by term.
 
  • #5
Tom Mattson said:
That should be [itex]t^{5/2}-\frac{1}{2}t^{3/2}[/itex] (you're missing the [itex]t[/itex] in the second term).



Why? You don't have a product anymore! Just differentiate term by term.

What do i do then? ;s

I'm stuck with t^(5/2) - (1/2)t^(3/2)
 
  • #6
Draggu said:
What do i do then? ;s

Just differentiate term by term.

Surely you know how to differentiate [itex]at^n[/itex] with respect to [itex]t[/itex], right? :confused:
 

What is the derivative of the given function?

The derivative of t^2[2\sqrt{t} - 1/\sqrt{t}] is 4t^{3/2} - 3/t^{3/2}.

How do you find the derivative of a function with multiple terms?

To find the derivative of a function with multiple terms, you can use the power rule and the sum rule. The power rule states that the derivative of a function raised to a constant power is equal to the constant times the function raised to the power minus one. The sum rule states that the derivative of a sum of functions is equal to the sum of the derivatives of each individual function.

What is the significance of the derivative in calculus?

The derivative is a fundamental concept in calculus and is used to measure the rate of change of a function. It represents the slope of a tangent line at a specific point on the function, and can be used to find maximum and minimum values, as well as determine the concavity of a 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. Similarly, a positive derivative indicates that the function is increasing at that point.

Can the derivative of a function be undefined?

Yes, the derivative of a function can be undefined at certain points where the function is not differentiable. This can happen at points where there is a sharp corner or cusp in the graph of the function, or at points where the function is not continuous.

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