# Derivative of this function please?

• nesan
In summary: Just remember that you can always cancel out x^-2 and x^-3, so you can make a common denominator by multiplying both terms by x.I'm sorry, I lost you. Could you please write out the steps you took so I can understand better?Sure, no problem. So far, I've done this:\frac{d}{dx}\frac{1+\sqrt{x}}{\sqrt[3]{x^2}}^3\\= \frac{d}{dx}\frac{(1 + \sqrt{x})^3}{x^2}\\= \frac{d}{dx}(1+\sqrt{x})^3x^{-2}To get to the last line, I used the
nesan

## Homework Statement

**NOTE**
We are not allowed to use the quotient rule. :"(

The function is

y = ( (1 + square-root(x)) / (cube-root(x^2)) )^3

## The Attempt at a Solution

I came up with a "simplified" expression.

(1 + x^0.5)^3 * (x^-2)

***This is what I did, but it seems very wrong.***

3(1 + x^.5)^2 * (0.5x^-0.5) + (1 + x^0.5)^3 * (-2x^-3)

When trying to simply that I got into a mess. :(

[3(1 + x^0.5)^2] / (2 * x^0.5) + [(1 + x^0.5)^3 * (-2x^-3)] / x^3

I'm stuck now. :(

nesan said:

## Homework Statement

**NOTE**
We are not allowed to use the quotient rule. :"(

The function is

y = ( (1 + square-root(x)) / (cube-root(x^2)) )^3

## The Attempt at a Solution

I came up with a "simplified" expression.

(1 + x^0.5)^3 * (x^-2)

***This is what I did, but it seems very wrong.***

3(1 + x^.5)^2 * (0.5x^-0.5) + (1 + x^0.5)^3 * (-2x^-3)
You're missing a factor here.
d/dx(f(x) * g(x)) = f'(x) * g(x) + g'(x) * f(x)
In your work you used the chain rule correctly to get f'(x), but you omitted g(x). Your other term looks fine, since you have both g'(x) and f(x).
nesan said:
When trying to simply that I got into a mess. :(

[3(1 + x^0.5)^2] / (2 * x^0.5) + [(1 + x^0.5)^3 * (-2x^-3)] / x^3

I'm stuck now. :(

nesan said:

## Homework Statement

**NOTE**
We are not allowed to use the quotient rule. :"(

The function is

y = ( (1 + square-root(x)) / (cube-root(x^2)) )^3
I guess this means$$\left ( \frac{1+\sqrt x}{x^{\frac 2 3}}\right)^3$$

## The Attempt at a Solution

I came up with a "simplified" expression.

(1 + x^0.5)^3 * (x^-2)

***This is what I did, but it seems very wrong.***

3(1 + x^.5)^2 * (0.5x^-0.5)x^-2 + (1 + x^0.5)^3 * (-2x^-3)

When trying to simply that I got into a mess. :(

[3(1 + x^0.5)^2] / (2 * x^0.5) + [(1 + x^0.5)^3 * (-2x^-3)] / x^3

I'm stuck now. :(

I added something you left out in red. You might not get a simple answer. You could maybe factor out some common factors.

In the line where you apply the Product Rule, check your first term--there's a factor missing. Once that's fixed, you should be able to multiply one of the terms by something to get them both over the same denominator.

You may also find it enlightening to try deriving the Quotient Rule in abstract form, that is:

$$\frac{d}{dx}\frac{u}{v} = \frac{d}{dx}(uv^{-1}) = ...$$

That's a bit easier to work with, because there aren't as many exponents and stuff as your real problem. But if you're able to figure out how to do that in abstract form, you can follow the same procedure to manipulate your actual expression.

Thanks guys, I'm working on it but I have a question.

Is it possible for me to cancel out the cube-root and the cube and only apply the cube to the numerator of the fraction only and leave the bottom as x^2?

Sure. That part of your simplification is fine, it's just the next step where things start going awry.

>_<

Lost again. :S

Here's what I have so far.

3(1 + x^.5)^2 * (0.5x^-0.5)x^-2 + (1 + x^0.5)^3 * (-2x^-3)

I moved everything with a negative exponent down to simplify.

3(1 + x^.5)^2 * (0.5x^-0.5)x^-2 + (1 + x^0.5)^3 * (-2x^-3)

Ended up with.

[3 * (1 + x^0.5)^2 * 0.5] / (x^0.5 * x^2) - (2 * (1 + x^0.5)^3) / x^3

Do I have to find a common denominator?

That seems scary. :(

Ok. So far, we have:

$$\frac{d}{dx}(\frac{1+\sqrt{x}}{\sqrt[3]{x^2}})^3\\ = \frac{d}{dx}\frac{(1 + \sqrt{x})^3}{x^2}\\ = \frac{d}{dx}(1+\sqrt{x})^3x^{-2}\\ = 3(1+\sqrt{x})^2\frac{1}{2\sqrt{x}x^2} +(1+\sqrt{x})^3 \frac{-2}{x^3}$$

You're probably not going to be able to get it a whole lot simpler than that, but you should at least be able to factor out few common things, and then get the rest over a common denominator.

## 1. What is a derivative?

A derivative is a mathematical concept that measures the rate of change of a function with respect to its independent variable. In other words, it tells us how much a function is changing at a specific point.

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

To find the derivative of a function, you need to use the rules of differentiation. These rules include the power rule, product rule, quotient rule, and chain rule. The specific rule used depends on the form of the function.

## 3. 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 slope of a tangent line, determine the maximum and minimum values of a function, and solve optimization problems.

## 4. How do you interpret the derivative of a function?

The derivative of a function can be interpreted as the instantaneous rate of change of the function at a specific point. It also tells us the direction in which the function is increasing or decreasing at that point.

## 5. 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 means the function is increasing, and a derivative of zero means the function is neither increasing nor decreasing.

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