#### Air conditionner

##### New member

- Joined
- Jun 29, 2021

- Messages
- 16

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Air conditionner
- Start date

- Joined
- Jun 29, 2021

- Messages
- 16

- Joined
- Jun 18, 2007

- Messages
- 24,973

It is not necessary to "separate" any variable to differentiate. Is the 't' independent variable? What is the dependent variable? "What" you have posted is not an equation - it does NOT have "equal" to ( = ) sign.Hi,

I have stumbled upon a little equation: I can't differentiate it.

Could anyon help me?

View attachment 28178

It's this one. I know how to deal with the products, and transform the root in fractional exponents, but I can't separate the t from the rest of the expression.

Thanks in advance,

Air conditioner

Why do you want to differentiate this expression - w.r.t. which variable?

Please post the EXACT problem as it was given to you.

Please show us what you have tried and

Please follow the rules of posting in this forum, as enunciated at:

Please share your work/thoughts about this problem

- Joined
- Nov 12, 2017

- Messages
- 11,778

This is a function of t, so presumably you want to differentiate it with respect to t. I don't know what you mean by "separate the t from the rest of the expression".Hi,

I have stumbled upon a little equation: I can't differentiate it.

Could anyon help me?

View attachment 28178

It's this one. I know how to deal with the products, and transform the root in fractional exponents, but I can't separate the t from the rest of the expression.

Thanks in advance,

Air conditioner

Please show whatever you have tried, including "transform the root into fractional exponents", so we can help with the next step.

- Joined
- Jun 29, 2021

- Messages
- 16

I want to find the derivative of y with respect to t. I know that 0.5 disappears since it's a constant. And after that, I can get this:

But I can't go further... I don't know how to expand with fractional exponents. Should I use the binomial theorem? I used an application (Gauthmath) to try to get an answer, and they did give a solution (where they didn't use the binomial theorem) but I didn't understand the steps...

- Joined
- Nov 12, 2017

- Messages
- 11,778

Do you know the chain rule? If `u = t - 125`, then this is `1/10 u^{1/3}`, which you should be able to differentiate.Well, this is the exercise: View attachment 28179

I want to find the derivative of y with respect to t. I know that 0.5 disappears since it's a constant. And after that, I can get this:

View attachment 28180

But I can't go further... I don't know how to expand with fractional exponents. Should I use the binomial theorem? I used an application (Gauthmath) to try to get an answer, and they did give a solution (where they didn't use the binomial theorem) but I didn't understand the steps...

- Joined
- Jun 29, 2021

- Messages
- 16

Thank you very much.

I am gonna search and learn the chain rule!

I am gonna search and learn the chain rule!

- Joined
- Jun 18, 2007

- Messages
- 24,973

\(\displaystyle y \ = \ 0.5 \ + \ \sqrt[3]{t \ - \ 125} \).............. substitute \(\displaystyle u \ = \ -125 \ + \ t \).........→...... \(\displaystyle \frac{du}{dt} \ = 1 \)Well, this is the exercise: View attachment 28179

I want to find the derivative of y with respect to t. I know that 0.5 disappears since it's a constant. And after that, I can get this:

View attachment 28180

But I can't go further... I don't know how to expand with fractional exponents. Should I use the binomial theorem? I used an application (Gauthmath) to try to get an answer, and they did give a solution (where they didn't use the binomial theorem) but I didn't understand the steps...

\(\displaystyle y \ = \ 0.5 \ + \ \sqrt[3]{u} \ = \ 0.5 + u^{\frac{1}{3}}\)

\(\displaystyle \frac{dy}{dt} \ = \frac{dy}{du} \ * \ \frac{du}{dt}\) .................. apply chain rule

\(\displaystyle \frac{dy}{dt} \ = \left[ \frac{1}{3}u^{(\frac{1}{3}-1)}\right] \ * \ \frac{du}{dt}\)

\(\displaystyle \frac{dy}{dt} \ = \left[ \frac{1}{3}u^{(-\frac{2}{3})}\right] \ * \ (1) \)

\(\displaystyle \frac{dy}{dt} \ = \ \frac{1}{3 \ * u^{(\frac{2}{3})}} \) ............. Now substitute back for "u".

- Joined
- Jun 29, 2021

- Messages
- 16

Thank you.\(\displaystyle y \ = \ 0.5 \ + \ \sqrt[3]{t \ - \ 125} \).............. substitute \(\displaystyle u \ = \ -125 \ + \ t \).........→...... \(\displaystyle \frac{du}{dt} \ = 1 \)

\(\displaystyle y \ = \ 0.5 \ + \ \sqrt[3]{u} \ = \ 0.5 + u^{\frac{1}{3}}\)

\(\displaystyle \frac{dy}{dt} \ = \frac{dy}{du} \ * \ \frac{du}{dt}\) .................. apply chain rule

\(\displaystyle \frac{dy}{dt} \ = \left[ \frac{1}{3}u^{(\frac{1}{3}-1)}\right] \ * \ \frac{du}{dt}\)

\(\displaystyle \frac{dy}{dt} \ = \left[ \frac{1}{3}u^{(-\frac{2}{3})}\right] \ * \ (1) \)

\(\displaystyle \frac{dy}{dt} \ = \ \frac{1}{3 \ * u^{(\frac{2}{3})}} \) ............. Now substitute back for "u".