Differentiate

[Isabella]

I am having trouble getting started with this one.

Differentiate: y=(x/(sqrt of 5-x))^3

 

[mmm4444bot]

I am having trouble getting started with this one.

Differentiate: y=(x/(sqrt of 5-x))^3

The phrase in red is not clear.

Is the function

y = (x/sqrt(5-x))^3

or

y = (x/(sqrt(5) - x))^3

or

something else? :cool:


EDIT: Please read the summary page of our posting guidelines HERE. Thank you.

 

[pka]

I am having trouble getting started with this one.

Differentiate: y=(x/(sqrt of 5-x))^3

Because the is no space in 5-x, I am reasonably sure it is: \(\displaystyle \left( {\dfrac{x}{{\sqrt {5 - x} }}} \right)\)

So here is a start: \(\displaystyle \displaystyle\dfrac{d}{{dx}}\left( {\dfrac{x}{{\sqrt {5 - x} }}} \right) = \frac{{\sqrt {5 - x} + \dfrac{x}{{2\sqrt {5 - x} }}}}{{5 - x}}\)

 

[irenevecchio]

Because the is no space in 5-x, I am reasonably sure it is: \(\displaystyle \left( {\dfrac{x}{{\sqrt {5 - x} }}} \right)\)

So here is a start: \(\displaystyle \displaystyle\dfrac{d}{{dx}}\left( {\dfrac{x}{{\sqrt {5 - x} }}} \right) = \frac{{\sqrt {5 - x} + \dfrac{x}{{2\sqrt {5 - x} }}}}{{5 - x}}\)

Thanks for the help. The problem started out correctly, but it is all to the power of 3.

y=(x/(sqrt of 5-x))^3 I am not sure how to start this.

 

[JeffM]

Thanks for the help. The problem started out correctly, but it is all to the power of 3.

y=(x/(sqrt of 5-x))^3 I am not sure how to start this.

Have you been taught the chain rule? More importantly, have you been taught how to USE the chain rule to simplify problems?

I see a cube. So \(\displaystyle let\ w = \dfrac{x}{\sqrt{5-x}} \implies y = w^3.\)

What is \(\displaystyle \dfrac{dy}{dw}?\)

Why is that relevant?

Now let's look at \(\displaystyle w = \dfrac{x}{\sqrt{5 - x}}.\)

I personally find working with products easier than working with quotients so I would restate as

\(\displaystyle w = x\left(\sqrt{5 - x}\right)^{(-1)}\).

Now that looks a bit ugly, but the multiplication rule tells me that if u and v are both differentiable functions of x and w = uv then

\(\displaystyle \dfrac{dw}{dx} = what?\)

That's not too hard so I let \(\displaystyle u= x\ and\ v = \left(\sqrt{5 - x}\right)^{(-1)}\).

Can you differentiate those two?

The second is still a little complex so let us know if you have trouble with it.

This may seem to be going nowhere, but in fact it is how you differentiate complicated functions. People who are really good at calculus do it in their heads, but if you are starting out, it helps to go step by step. We are happy to walk you through it, but to learn it, you need to do it.

 

[irenevecchio]

Have you been taught the chain rule? More importantly, have you been taught how to USE the chain rule to simplify problems?

I see a cube. So \(\displaystyle let\ w = \dfrac{x}{\sqrt{5-x}} \implies y = w^3.\)

What is \(\displaystyle \dfrac{dy}{dw}?\)

Why is that relevant?

Now let's look at \(\displaystyle w = \dfrac{x}{\sqrt{5 - x}}.\)

I personally find working with products easier than working with quotients so I would restate as

\(\displaystyle w = x\left(\sqrt{5 - x}\right)^{(-1)}\).


Now that looks a bit ugly, but the multiplication rule tells me that if u and v are both differentiable functions of x and w = uv then

\(\displaystyle \dfrac{dw}{dx} = what?\)

That's not too hard so I let \(\displaystyle u= x\ and\ v = \left(\sqrt{5 - x}\right)^{(-1)}\).

Can you differentiate those two?

The second is still a little complex so let us know if you have trouble with it.

This may seem to be going nowhere, but in fact it is how you differentiate complicated functions. People who are really good at calculus do it in their heads, but if you are starting out, it helps to go step by step. We are happy to walk you through it, but to learn it, you need to do it.

Ok. Tell me if I am starting this correctly? Not quite sure if I am doing the power correctly.

y=(x/(sqrt5-x))^3
=3(x/(5-x)^1/2)^2
=3(x(5-x))^3/2
=3(x^3/2(5^3/2-x^3/2)

 

[Isabella]

Ok. Tell me if I am starting this correctly? Not quite sure if I am doing the power correctly.

y=(x/(sqrt5-x))^3
=3(x/(5-x)^1/2)^2
=3(x(5-x))^3/2
=3(x^3/2(5^3/2-x^3/2)

I appreciate your help JeffM & irenevecchio. I am having a time with starting this same problem out. I am not sure if you have the powers correctly, that is where I am having the hardest time.

Please continue to work this. I totally do not understand!

 

[JeffM]

Ok. Tell me if I am starting this correctly? Not quite sure if I am doing the power correctly.

y=(x/(sqrt5-x))^3
=3(x/(5-x)^1/2)^2
=3(x(5-x))^3/2
=3(x^3/2(5^3/2-x^3/2)

You sort of started to get what I was driving at, but your notation is not helping you (in fact it is wrong) so let's start over. This is easier to comprehend if you use the very old fashioned notation for derivatives of dy/dx and so on.

Finding the derivative, which is just a function, of a function that is complicated involves breaking the complicated function into simpler functions and applying a handful of rules for finding the derivatives of simpler functions. This breaking a function down into component parts that are easy to deal with may involve doing so repeatedly.

\(\displaystyle y = \left(\dfrac{x}{\sqrt{x - 5}}\right)^3\ is\ given.\ We\ want\ to\ find\ \dfrac{dy}{dx}.\)

So my first idea for simplification is to substitute a single variable for what is being cubed to get a nice simple function.

\(\displaystyle w = \dfrac{x}{\sqrt{x - 5}} \implies y = w^3.\) Do you have any confusion about this step?

Now I can apply the POWER RULE.

\(\displaystyle y = w^3 \implies \dfrac{dy}{dw} = 3w^2 = 3\left(\dfrac{x}{\sqrt{x - 5}}\right)^2.\)

You saw this, sort of, but note that what you wrote was "y = (x/(sqrt5-x))^3 = 3(x/(5-x)^1/2)^2." That is NOT true. The line above is what is true. If you do not understand this, let me know because you need to understand each step.

However, you are not trying to find dy/dw. Rather you are trying to find dy/dx. This leads us to the CHAIN RULE, a VERY important rule.

\(\displaystyle y = w^3 = \left(\dfrac{x}{\sqrt{x - 5}}\right)^3\) means that y is function of w, which is a function of x. That in turn means that y is

a so-called composite function of x, and the chain rule explains how to find the derivative of a composite function.

\(\displaystyle y = w^3 = \left(\dfrac{x}{\sqrt{x - 5}}\right)^3 \implies \dfrac{dy}{dx} = \dfrac{dy}{dw} * \dfrac{dw}{dx}.\) And we know what \(\displaystyle \dfrac{dy}{dw}\) is.

I find it easy to remember the chain rule in this form because it looks as though the dw terms cancel out like fractions (although this is not correct if we are being rigorous).

So putting together what we have this far: \(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{dw} * \dfrac{dw}{dx} = 3w^2 * \dfrac{dw}{dx } = 3\left(\dfrac{x}{\sqrt{x - 5}}\right)^2 * \dfrac{dw}{dx}.\)

The next step is to find \(\displaystyle \dfrac{dw}{dx}.\) Are you with me to here?

 

[Isabella]

You sort of started to get what I was driving at, but your notation is not helping you (in fact it is wrong) so let's start over. This is easier to comprehend if you use the very old fashioned notation for derivatives of dy/dx and so on.

Finding the derivative, which is just a function, of a function that is complicated involves breaking the complicated function into simpler functions and applying a handful of rules for finding the derivatives of simpler functions. This breaking a function down into component parts that are easy to deal with may involve doing so repeatedly.

\(\displaystyle y = \left(\dfrac{x}{\sqrt{x - 5}}\right)^3\ is\ given.\ We\ want\ to\ find\ \dfrac{dy}{dx}.\)

So my first idea for simplification is to substitute a single variable for what is being cubed to get a nice simple function.

\(\displaystyle w = \dfrac{x}{\sqrt{x - 5}} \implies y = w^3.\) Do you have any confusion about this step?

Now I can apply the POWER RULE.

\(\displaystyle y = w^3 \implies \dfrac{dy}{dw} = 3w^2 = 3\left(\dfrac{x}{\sqrt{x - 5}}\right)^2.\)

You saw this, sort of, but note that what you wrote was "y = (x/(sqrt5-x))^3 = 3(x/(5-x)^1/2)^2." That is NOT true. The line above is what is true. If you do not understand this, let me know because you need to understand each step.

However, you are not trying to find dy/dw. Rather you are trying to find dy/dx. This leads us to the CHAIN RULE, a VERY important rule.

\(\displaystyle y = w^3 = \left(\dfrac{x}{\sqrt{x - 5}}\right)^3\) means that y is function of w, which is a function of x. That in turn means that y is

a so-called composite function of x, and the chain rule explains how to find the derivative of a composite function.

\(\displaystyle y = w^3 = \left(\dfrac{x}{\sqrt{x - 5}}\right)^3 \implies \dfrac{dy}{dx} = \dfrac{dy}{dw} * \dfrac{dw}{dx}.\) And we know what \(\displaystyle \dfrac{dy}{dw}\) is.

I find it easy to remember the chain rule in this form because it looks as though the dw terms cancel out like fractions (although this is not correct if we are being rigorous).

So putting together what we have this far: \(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{dw} * \dfrac{dw}{dx} = 3w^2 * \dfrac{dw}{dx } = 3\left(\dfrac{x}{\sqrt{x - 5}}\right)^2 * \dfrac{dw}{dx}.\)

The next step is to find \(\displaystyle \dfrac{dw}{dx}.\) Are you with me to here?

So then 3(x/(sqrt 5-x)^2 * (x/(sqrt 5-x) ?

 

[JeffM]

So then 3(x/(sqrt 5-x)^2 * (x/(sqrt 5-x) ?

No. What you have there is \(\displaystyle \dfrac{dy}{dw} * w \ne \dfrac{dy}{dw} * \dfrac{dw}{dx}.\)

You have to distinguish between functions and their derivatives.

\(\displaystyle w = \dfrac{x}{\sqrt{5 - x}} = x\left(\sqrt{5 - x}\right)^{(-1)} \implies \dfrac{dw}{dx} = what?\)

This is another complex function so what do you do? Do you know a rule that relates to products? What does that rule suggest doing next?

 

[Quaid]

having trouble getting started

Differentiate: y = [x/sqrt(5-x)]^3

Hi Isabella:

Here is a different approach. Express y as a product.

y = x^3 * (5 - x)^(-3/2)

The product rule may be used, to find y'

(5 - x)^(-3/2) is a composite function; you'll need to apply the chain rule, when calculating its derivative.

Cheers :cool:

 

[Isabella]

Hi Isabella:

Here is a different approach. Express y as a product.

y = x^3 * (5 - x)^(-3/2)

The product rule may be used, to find y'

(5 - x)^(-3/2) is a composite function; you'll need to apply the chain rule, when calculating its derivative.

Cheers :cool:

Thanks Quaid. All help is very appreciated.