Understanding the Chain Rule in Multivariable Calculus

[Physics345]

Homework Statement

Question: FIND $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial t}$

Given

$z= x^{y}$

$x =\sqrt{s+t}$

$y=ts^{2}$

Relevant Equations

$\frac{\partial z}{\partial t} =\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$

Solution:
$\frac{\partial z}{\partial x} = yx^{y -1}+1$

$\frac{\partial z}{\partial t} =\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$
$\frac{\partial z}{\partial t} = (\frac{yx^{y-1} + 1}{2\sqrt{s+t}}) + x^{y}\ln{(y)}s^{2}$

This is my first time doing the chain rule and my professor doesn't give us the answers for questions. So I'm here to ask you guys if I did this correct. I'm very confused was I supposed to input the values given after completion or not?

Thanks

 

[PeroK]

Question: dz/dx, dz/dt where z = x^y + x x =sqrt(s+t) y=ts^2

dz/dx = yx^y + 1

dz/dt = dz/dx * dx/dt + dz/dy * dy/dt

dz/dt = (yx^(y-1) + 1)*(1/(2sqrt(s+t)) + ((x^y)ln(y))*(s^2)

= yx^(y-1)/(2sqrt(s+t)) +2sqrt(s+t) + ln(y)*(sx^y)

This is my first time doing the chain rule and my professor doesn't give us the answers for questions. So I'm here to ask you guys if I did this correct. I'm very confused was I supposed to input the values given after completion or not?

Thanks

You should learn a bit of Latex:

https://www.physicsforums.com/help/latexhelp/

​

I find it almost impossible to parse what you've written but you have the right idea.​

 

[Physics345]

You should learn a bit of Latex:

https://www.physicsforums.com/help/latexhelp/

​

I find it almost impossible to parse what you've written but you have the right idea.​

Thanks, I have bookmarked it. Give me a moment to apply the latex.

 

[PeroK]

Question: dz/dx, dz/dt where z = x^y + x x =sqrt(s+t) y=ts^2

dz/dx = yx^y + 1

dz/dt = dz/dx * dx/dt + dz/dy * dy/dt

dz/dt = (yx^(y-1) + 1)*(1/(2sqrt(s+t)) + ((x^y)ln(y))*(s^2)

= yx^(y-1)/(2sqrt(s+t)) +2sqrt(s+t) + ln(y)*(sx^y)

This is my first time doing the chain rule and my professor doesn't give us the answers for questions. So I'm here to ask you guys if I did this correct. I'm very confused was I supposed to input the values given after completion or not?

Thanks

To give you start with Latex:

$\frac{\partial z}{\partial t} = (yx^{y-1} + 1)(\frac{1}{2\sqrt{s+t}}) + \dots$

If you reply to this, you can pick up my Latex to edit or cut and paste.

 

[Physics345]

To give you start with Latex:

$\frac{\partial z}{\partial t} = (yx^{y-1} + 1)(\frac{1}{2\sqrt{s+t}}) + \dots$

If you reply to this, you can pick up my Latex to edit or cut and paste.

Done thanks a lot.

 

[PeroK]

Done thanks a lot.

For your final answer, you should replace $x, y$ with the relevant functions of $s, t$. Also, you have $\ln y$ instead of $\ln x$.

You've definitely got the right approach.

 

[SammyS]

Solution:
$\frac{\partial z}{\partial x} = yx^{y -1}+1$

That doesn't look right to me.

Where does the 1 come from and why are you adding it?

 

[Physics345]

That doesn't look right to me.

Where does the 1 come from and why are you adding it?

Oh, I just realized that. I must have added it as I was applying the latex.

 

[Physics345]

For your final answer, you should replace $x, y$ with the relevant functions of $s, t$. Also, you have $\ln y$ instead of $\ln x$.

You've definitely got the right approach.

Another typo. I wish I saw your message earlier. I had my exam today and did not substitute the values.

 

[Physics345]

Either way, the concept was simple. I will come back again when I'm confused.

Thanks, Everyone & Best Wishes,

Physics345