Quotient rule + simplifying the result

[BOAS]

Homework Statement

Hello,

I have an answer to my question but I'm not sure if I've simplified it correctly or if it can boil down further.

differentiate;

y = \frac{z^{3}-z}{\sqrt{z}}

Homework Equations

Quotient rule

The Attempt at a Solution

let u = z³ - z
let v = z^1/2

du/dx = 3z² - 1

dv/dx = 1/2 z^-1/2

dy/dx = \frac{z^{1/2}(3z^{2}-1) - (z^{3}-z)1/2z^{-1/2}}{(z^{1/2})^{2}}


dy/dx = \frac{((3z^{2})(z^{1/2}) - z^{1/2}) - (z^{3})(1/2z^{-1/2})-z(z^{-1/2})}{z}

dy/dx = \frac{(3z^{5/2} - z^{1/2}) - (1/2z^{5/2} - z^{1/2})}{z}

dy/dx = \frac{5/2 z^{5/2}}{z}

Thanks!

 

[LCKurtz]

Homework Statement

Hello,

I have an answer to my question but I'm not sure if I've simplified it correctly or if it can boil down further.

differentiate;

y = \frac{z^{3}-z}{\sqrt{z}}

Homework Equations

Quotient rule

The Attempt at a Solution

let u = z³ - z
let v = z^1/2

du/dx = 3z² - 1

dv/dx = 1/2 z^-1/2

dy/dx = \frac{z^{1/2}(3z^{2}-1) - (z^{3}-z)1/2z^{-1/2}}{(z^{1/2})^{2}}

It isn't dy/dx you are calculating. It is dy/dz. After the above step, the next thing you should do is multiply both numerator and denominator by $2\sqrt z$. That will make the remaining steps much easier.

dy/dx = \frac{((3z^{2})(z^{1/2}) - z^{1/2}) - (z^{3})(1/2z^{-1/2})-z(z^{-1/2})}{z}

dy/dx = \frac{(3z^{5/2} - z^{1/2}) - (1/2z^{5/2} - z^{1/2})}{z}

dy/dx = \frac{5/2 z^{5/2}}{z}

Thanks!

Your final answer is neither simplified nor correct.

 

[BOAS]

It isn't dy/dx you are calculating. It is dy/dz.

Woops, Thanks for catching that.

After the above step, the next thing you should do is multiply both numerator and denominator by $2\sqrt z$. That will make the remaining steps much easier.

Well, you've identified a weak spot for me. Surds.

I'm struggling with the top line, I think the bottom line looks like this;

dy/dz = \frac{2\sqrt{z}(z^{1/2}(3z^{2}-1) - 1/2z^{-1/2}(z^{3} - z))}{z^{2}}

For the top line, do I need to multiply 2√z into the bits outside the innermost brackets?

 

[Mark44]

Woops, Thanks for catching that.
Well, you've identified a weak spot for me. Surds.

I'm struggling with the top line, I think the bottom line looks like this;

dy/dz = \frac{2\sqrt{z}(z^{1/2}(3z^{2}-1) - 1/2z^{-1/2}(z^{3} - z))}{z^{2}}

For the top line, do I need to multiply 2√z into the bits outside the innermost brackets?

It's probably simpler to do the operations that are in the outer-most set of parentheses, and combine like terms. Then you can multiply what's left by 2√z.

It's also better to work with fractional exponents here than with radicals. √z is the same as z^1/2. You'll confuse yourself if you have expressions with mixed radicals and exponents.

For this problem, it's easy to check your work by simplifying the function first, and then differentiating. You should get the same answer whether you use the quotient rule or some other rule.

In this case, y = $\frac{z^3 - z}{z^{1/2}} = z^{5/2} - z^{1/2}$. From this it's simple to get dy/dz.

 

[LCKurtz]

Woops, Thanks for catching that.



Well, you've identified a weak spot for me. Surds.

I'm struggling with the top line, I think the bottom line looks like this;

dy/dz = \frac{2\sqrt{z}(z^{1/2}(3z^{2}-1) - 1/2z^{-1/2}(z^{3} - z))}{z^{2}}

For the top line, do I need to multiply 2√z into the bits outside the innermost brackets?

You need to distribute the $2\sqrt z$ across both terms in the outer red parentheses$$
dy/dz = \frac{2\sqrt{z}\color{red}(z^{1/2}(3z^{2}-1) -
\color{red}(1/2\color{red})z^{-1/2}(z^{3} - z)\color{red})}{z^{2}}$$That will give you a $2z$ factor in the first term and will cancel the $\frac 1 2 z^{-\frac 1 2}$ in the second term of the numerator. Also check your exponent in the denominator.