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

• Draggu
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.

## Homework Statement

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

## 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')

Let's see what you did. Then we can point out where you're going wrong.

=(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 [($$\sqrt{t}$$(10t-3)/(4)]

Draggu said:
=(1/2)t^2[(2t^(1/2)) - (t^(-1/2))]
=t^(5/2) - (1/2)^(3/2)

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

Use product rule

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

Tom Mattson said:
That should be $t^{5/2}-\frac{1}{2}t^{3/2}$ (you're missing the $t$ 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)

Draggu said:
What do i do then? ;s

Just differentiate term by term.

Surely you know how to differentiate $at^n$ with respect to $t$, right?

## 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.